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# Lecture 11:Sinusoidal Steady-State (SSS) Or Frequency Response

Summary: SSS response essentially characterizes the system function. SSS response is commonly used to measure the system function. A common mode of thinking for signal processing tasks is “filtering.” All physical systems act as filters of their input signals. Filtering is an important signal processing method.

Lecture #11:

SINUSOIDAL STEADY-STATE (SSS) OR FREQUENCY RESPONSE

Motivation:

• Many systems operate in the SSS, e.g., electrical power distribution, broadcast, touch-tone telephone.
• SSS response essentially characterizes the system function.
• SSS response is commonly used to measure the system function.
• A common mode of thinking for signal processing tasks is “filtering.” All physical systems act as filters of their input signals. Filtering is an important signal processing method.

Outline:

• Causal, stable systems
• Sinusoidal steady-state response—the frequency response
• Relation of frequency response to system function
• Bode diagrams
• Signal processing with filters
• Lowpass and highpass filters
• Resonance and bandpass filters
• Notch filters
• Conclusions

A system that operates in the SSS . . . well almost

The touch-tone phone

Dialing consists of pressing buttons on the keypad which has 3 columns and 4 rows. How is the information about which button is pushed coded?

Demo on coding in the touch-tone phone.

Sum of sinusoids

s ( t ) = sin ( 2πf 1 t ) + sin ( 2πf 2 t ) s ( t ) = sin ( 2πf 1 t ) + sin ( 2πf 2 t ) size 12{s $$t$$ ="sin" $$2πf rSub { size 8{1} } t$$ +"sin" $$2πf rSub { size 8{2} } t$$ } {}

times a pulse window p(t)

x ( t ) = p ( t ) s ( t ) x ( t ) = p ( t ) s ( t ) size 12{x $$t$$ =p $$t$$ s $$t$$ } {}

More on the effect of p(t) later!

I. CAUSAL, STABLE SYSTEMS

The system function of an LTI system, H(s), can be used to categorize systems.

• Causal system size 12{ drarrow } {} ROC is to the right of the rightmost pole of H(s).
• Causal, stable system size 12{ drarrow } {} ROC is to the right of the rightmost pole of H(s) and all the poles are in the left-half of the s plane.

More on the definition of stable systems at a later time.

A causal, stable system has the pole-zero diagram shown below.

For s in the shaded region, the response to x(t)=Xestx(t)=Xest size 12{x $$t$$ = ital "Xe" rSup { size 8{ ital "st"} } } {} is the steady-state response y(t)=Yest=XH(s)esty(t)=Yest=XH(s)est size 12{y $$t$$ = ital "Ye" rSup { size 8{ ital "st"} } = ital "XH" $$s$$ e rSup { size 8{ ital "st"} } } {}

II. THE SYSTEM FUNCTION H(S)

1/ Real and complex poles

H(s) is a complex function of a complex variable s. The plots show |H(s)|.

2/ Effect of a zero

3/ Interpretation of H(s) by pole and zero vectors

H(s) that is a rational function has the form

H ( s ) = K ( s z 1 ) ( s z 2 ) . . . ( s z M ) ( s p 1 ) ( s p 2 ) . . . ( s p N ) H ( s ) = K ( s z 1 ) ( s z 2 ) . . . ( s z M ) ( s p 1 ) ( s p 2 ) . . . ( s p N ) size 12{H $$s$$ =K { { $$s - z rSub { size 8{1} }$$ $$s - z rSub { size 8{2} }$$ "." "." "." $$s - z rSub { size 8{M} }$$ } over { $$s - p rSub { size 8{1} }$$ $$s - p rSub { size 8{2} }$$ "." "." "." $$s - p rSub { size 8{N} }$$ } } } {}

H(s) consists of products and quotients of the form (ssk)(ssk) size 12{ $$s - s rSub { size 8{k} }$$ } {}. Each of these terms are vectors in the complex s plane:

(sz1)(sz2)...(szM)(sz1)(sz2)...(szM) size 12{ $$s - z rSub { size 8{1} }$$ $$s - z rSub { size 8{2} }$$ "." "." "." $$s - z rSub { size 8{M} }$$ } {} are called zero vectors,

(sp1)(sp2)...(spN)(sp1)(sp2)...(spN) size 12{ $$s - p rSub { size 8{1} }$$ $$s - p rSub { size 8{2} }$$ "." "." "." $$s - p rSub { size 8{N} }$$ } {}are called pole vectors.

=

The vector (ssk)(ssk) size 12{ $$s - s rSub { size 8{k} }$$ } {}points from sksk size 12{s rSub { size 8{k} } } {} to s. It can be expressed in polar form as

( s s k ) = s s k e j arg ( s s k ) ( s s k ) = s s k e j arg ( s s k ) size 12{ $$s - s rSub { size 8{k} }$$ = \lline s - s rSub { size 8{k} } \lline e rSup { size 8{j"arg" $$s - s rSub { size 6{k} }$$ } } } {}

We now take

H ( s ) = K ( s z 1 ) ( s z 2 ) . . . ( s z M ) ( s p 1 ) ( s p 2 ) . . . ( s p N ) H ( s ) = K ( s z 1 ) ( s z 2 ) . . . ( s z M ) ( s p 1 ) ( s p 2 ) . . . ( s p N ) size 12{H $$s$$ =K { { $$s - z rSub { size 8{1} }$$ $$s - z rSub { size 8{2} }$$ "." "." "." $$s - z rSub { size 8{M} }$$ } over { $$s - p rSub { size 8{1} }$$ $$s - p rSub { size 8{2} }$$ "." "." "." $$s - p rSub { size 8{N} }$$ } } } {}

and express the vectors in polar form

{} H ( s ) = K s z 1 e j arg ( s z 1 ) s z 2 e j arg ( s z 2 ) . . . s z M e j arg ( s z M ) s p 1 e j arg ( s p 1 ) s p 2 e j arg ( s p 2 ) . . . s p N e j arg ( s p N ) H ( s ) = K s z 1 e j arg ( s z 1 ) s z 2 e j arg ( s z 2 ) . . . s z M e j arg ( s z M ) s p 1 e j arg ( s p 1 ) s p 2 e j arg ( s p 2 ) . . . s p N e j arg ( s p N ) size 12{H $$s$$ =K { { \lline s - z rSub { size 8{1} } \lline e rSup { size 8{j"arg" $$s - z rSub { size 6{1} }$$ } } \lline s - z rSub {2} size 12{ \lline e rSup {j"arg" $$s - z rSub { size 6{2} }$$ } } size 12{ "." "." "." \lline s - z rSub {M} } size 12{ \lline e rSup {j"arg" $$s - z rSub { size 6{M} }$$ } }} over { size 12{ \lline s - p rSub {1} size 12{ \lline e rSup {j"arg" $$s - p rSub { size 6{1} }$$ } } size 12{ \lline s - p rSub {2} } size 12{ \lline e rSup {j"arg" $$s - p rSub { size 6{2} }$$ } } size 12{ "." "." "." \lline s - p rSub {N} } size 12{ \lline e rSup {j"arg" $$s - p rSub { size 6{N} }$$ } }} } } } {}

so that

H ( s ) = K s z 1 s z 2 . . . s z M s p 1 s p 2 . . . s p M H ( s ) = K s z 1 s z 2 . . . s z M s p 1 s p 2 . . . s p M size 12{ \lline H $$s$$ \lline = \lline K \lline { { \lline s - z rSub { size 8{1} } \lline \lline s - z rSub { size 8{2} } \lline "." "." "." \lline s - z rSub { size 8{M} } \lline } over { \lline s - p rSub { size 8{1} } \lline \lline s - p rSub { size 8{2} } \lline "." "." "." \lline s - p rSub { size 8{M} } \lline } } } {}

And

arg H ( s ) = arg K + arg ( s z 1 ) + arg ( s z 2 ) + . . . + arg ( s z M ) arg ( s p 1 ) arg ( s p 2 ) . . . arg ( s p N ) arg H ( s ) = arg K + arg ( s z 1 ) + arg ( s z 2 ) + . . . + arg ( s z M ) arg ( s p 1 ) arg ( s p 2 ) . . . arg ( s p N ) size 12{"arg"H $$s$$ ="arg"K+"arg" $$s - z rSub { size 8{1} }$$ +"arg" $$s - z rSub { size 8{2} }$$ + "." "." "." +"arg" $$s - z rSub { size 8{M} }$$ - "arg" $$s - p rSub { size 8{1} }$$ - "arg" $$s - p rSub { size 8{2} }$$ - "." "." "." - "arg" $$s - p rSub { size 8{N} }$$ } {}

And

H ( s ) = H ( s ) e j arg ( H ( s ) ) H ( s ) = H ( s ) e j arg ( H ( s ) ) size 12{H $$s$$ = \lline H $$s$$ \lline e rSup { size 8{j"arg" $$H \( s$$ \) } } } {}

III. FREQUENCY RESPONSE

1/ Relation to system function

Note if

x ( t ) = Xe jωt + Xe jωt 2 = X cos ( ωt ) x ( t ) = Xe jωt + Xe jωt 2 = X cos ( ωt ) size 12{x $$t$$ = { { ital "Xe" rSup { size 8{jωt} } + ital "Xe" rSup { size 8{ - jωt} } } over {2} } =X"cos" $$ωt$$ } {}

then

y ( t ) = XH ( ) e jωt + XH ( ) e jωt 2 y ( t ) = XH ( ) e jωt 2 + XH ( ) e jωt 2 y ( t ) = 2R XH ( ) e jωt 2 y ( t ) = 2R X H ( ) e j arg ( H ( ) ) e jωt y ( t ) = X H ( ) cos ( ωt + arg ( H ( ) ) ) y ( t ) = XH ( ) e jωt + XH ( ) e jωt 2 y ( t ) = XH ( ) e jωt 2 + XH ( ) e jωt 2 y ( t ) = 2R XH ( ) e jωt 2 y ( t ) = 2R X H ( ) e j arg ( H ( ) ) e jωt y ( t ) = X H ( ) cos ( ωt + arg ( H ( ) ) ) alignl { stack { size 12{y $$t$$ = { { ital "XH" $$jω$$ e rSup { size 8{jωt} } + ital "XH" $$- jω$$ e rSup { size 8{ - jωt} } } over {2} } } {} # y $$t$$ = { { ital "XH" $$jω$$ e rSup { size 8{jωt} } } over {2} } + left [ { { ital "XH" $$jω$$ e rSup { size 8{jωt} } } over {2} } right ] rSup { size 8{*} } {} # y $$t$$ =2R left lbrace { { ital "XH" $$jω$$ e rSup { size 8{jωt} } } over {2} } right rbrace {} # y $$t$$ =2R left lbrace X \lline H $$jω$$ \lline e rSup { size 8{j"arg" $$H \( jω$$ \) } } e rSup { size 8{jωt} } right rbrace {} # y $$t$$ =X \lline H $$jω$$ \lline "cos" $$ωt+"arg" \( H \( jω$$ \) \) {} } } {}

2/ H(s) along the imaginary axis

H(s) evaluated for s = jω or H(jω) is called the frequency response.

• ω can be expressed as ω = 2πf.
• f is the frequency in units of cycles per second which is called hertz (Hz).
• The frequency response H(j2πf) is often plotted versus f.

4/ Measurement of H(jω)

To measure H(jω) of a test system we can use the system shown below.

However, when the sinusoid is turned on the response contains both a transient and a steady-state component.

For example, suppose

H ( s ) = 5s ( s + 1 ) 2 + 10 2 H ( s ) = 5s ( s + 1 ) 2 + 10 2 size 12{H $$s$$ = { {5s} over { $$s+1$$ rSup { size 8{2} } +"10" rSup { size 8{2} } } } } {}

The response of this system, y(t), to the input x(t) = cos(2πt) u(t) is shown below.

Note the transient at the onset which damps out after a few cycles of the sinusoid so that the response approaches the particular solution, i.e., this is the steady-state response.

Therefore, to measure H(jω) we turn on the oscillator and wait till steady state is established and measure the input and output sinusoid. At each frequency, the ratio of the magnitude of the output to the magnitude of the input sinusoid defines the magnitude of the frequency response. The angle of the output minus that of the input defines the angle of the frequency response.

More elaborate systems are available for measuring the frequency response rapidly and automatically.

5/ Relation of time waveforms, vector diagrams, and frequency response

Consider and LTI system with system function

H ( s ) = 1 s + 1 H ( s ) = 1 s + 1 size 12{H $$s$$ = { {1} over {s+1} } } {}

which has the frequency response

H ( ) = 1 + 1 H ( ) = 1 + 1 size 12{H $$jω$$ = { {1} over {jω+1} } } {}

The input is

x ( t ) cos ( ωt ) x ( t ) cos ( ωt ) size 12{x $$t$$ "cos" $$ωt$$ } {}

and the response is

y ( t ) = H ( ) cos ( ωt + arg H ( ) ) y ( t ) = 1 ( ω 2 + 1 ) 1 / 2 cos ( ωt tan 1 ω ) y ( t ) = H ( ) cos ( ωt + arg H ( ) ) y ( t ) = 1 ( ω 2 + 1 ) 1 / 2 cos ( ωt tan 1 ω ) alignl { stack { size 12{y $$t$$ = \lline H $$jω$$ \lline "cos" $$ωt+"arg"H \( jω$$ \) } {} # size 12{y $$t$$ = { {1} over { $$ω rSup { size 8{2} } +1$$ rSup { size 8{1/2} } } } "cos" $$ωt - "tan" rSup { size 8{ - 1} } ω$$ } {} } } {}

Demo of relation of pole-zero diagram, time waveforms, vector diagrams, and frequency response.

6/ Different ways of plotting the frequency response

The frequency response H(jω) = 1/(jω +1), is plotted in linear coordinates (left) and in doubly logarithmic coordinates called a Bode diagram (right).

IV. BODE DIAGRAMS

1/ Definition and rationale

Frequency responses are commonly plotted as Bode diagrams which are plots of

20log10H()20log10H() size 12{"20""log" rSub { size 8{"10"} } \lline H $$jω$$ \lline } {} plotted versus log10ωlog10ω size 12{"log" rSub { size 8{"10"} } ω} {}

argH()argH() size 12{"arg"H $$jω$$ } {} plotted versus log10ωlog10ω size 12{"log" rSub { size 8{"10"} } ω} {}.

The reasons are:

• Logarithmic coordinates are useful when the range of |H(jω)| and/or ω is large.
• Asymptotes to the frequency response are easily plotted.
• When the poles and zeros are on the real axis, the asymptotes are excellent approximations to the frequency response.

2/ Pole-zero and time-constant form

Consider an LTI system whose system function has poles and zeros on the negative, real axis. It can be displayed in pole-zero form as follows

H ( s ) = C ( s + z 1 ) ( s + z 2 ) . . . ( s + z M ) ( s + p 1 ) ( s + p 2 ) . . . ( s + p N ) H ( s ) = C ( s + z 1 ) ( s + z 2 ) . . . ( s + z M ) ( s + p 1 ) ( s + p 2 ) . . . ( s + p N ) size 12{H $$s$$ =C { { $$s+z rSub { size 8{1} }$$ $$s+z rSub { size 8{2} }$$ "." "." "." $$s+z rSub { size 8{M} }$$ } over { $$s+p rSub { size 8{1} }$$ $$s+p rSub { size 8{2} }$$ "." "." "." $$s+p rSub { size 8{N} }$$ } } } {}

This system function can be evaluated along the jω axis to yield

H ( s ) = C ( + z 1 ) ( + z 2 ) . . . ( + z M ) ( + p 1 ) ( + p 2 ) . . . ( + p N ) H ( s ) = C ( + z 1 ) ( + z 2 ) . . . ( + z M ) ( + p 1 ) ( + p 2 ) . . . ( + p N ) size 12{H $$s$$ =C { { $$jω+z rSub { size 8{1} }$$ $$jω+z rSub { size 8{2} }$$ "." "." "." $$jω+z rSub { size 8{M} }$$ } over { $$jω+p rSub { size 8{1} }$$ $$jω+p rSub { size 8{2} }$$ "." "." "." $$jω+p rSub { size 8{N} }$$ } } } {}

By dividing the numerator by each zero and the denominator by each pole, the frequency response can be put in time-constant form as follows

H ( ) = K ( j ωτ z1 + 1 ) ( j ωτ z2 + 1 ) . . . ( j ωτ zM + 1 ) ( j ωτ p1 + 1 ) ( j ωτ p2 + 1 ) . . . ( j ωτ pN + 1 ) H ( ) = K ( j ωτ z1 + 1 ) ( j ωτ z2 + 1 ) . . . ( j ωτ zM + 1 ) ( j ωτ p1 + 1 ) ( j ωτ p2 + 1 ) . . . ( j ωτ pN + 1 ) size 12{H $$jω$$ =K { { $$j ital "ωτ" rSub { size 8{z1} } +1$$ $$j ital "ωτ" rSub { size 8{z2} } +1$$ "." "." "." $$j ital "ωτ" rSub { size 8{ ital "zM"} } +1$$ } over { $$j ital "ωτ" rSub { size 8{p1} } +1$$ $$j ital "ωτ" rSub { size 8{p2} } +1$$ "." "." "." $$j ital "ωτ" rSub { size 8{ ital "pN"} } +1$$ } } } {}

3/ Magnitude and angle

The magnitude and angle can be expressed as

H ( ) = K j ωτ z1 + 1 j ωτ z2 + 1 . . . j ωτ zM + 1 j ωτ p1 + 1 j ωτ p2 + 1 . . . j ωτ pN + 1 H ( ) = K j ωτ z1 + 1 j ωτ z2 + 1 . . . j ωτ zM + 1 j ωτ p1 + 1 j ωτ p2 + 1 . . . j ωτ pN + 1 size 12{ \lline H $$jω$$ \lline = \lline K \lline { { \lline j ital "ωτ" rSub { size 8{z1} } +1 \lline \lline j ital "ωτ" rSub { size 8{z2} } +1 \lline "." "." "." \lline j ital "ωτ" rSub { size 8{ ital "zM"} } +1 \lline } over { \lline j ital "ωτ" rSub { size 8{p1} } +1 \lline \lline j ital "ωτ" rSub { size 8{p2} } +1 \lline "." "." "." \lline j ital "ωτ" rSub { size 8{ ital "pN"} } +1 \lline } } } {}

And

arg ( H ( ) ) = arg ( K ) + arg ( j ωτ z1 + 1 ) + arg ( j ωτ z2 + 1 ) + . . . + arg ( j ωτ zM + 1 ) arg ( j ωτ p1 + 1 ) arg ( j ωτ p2 + 1 ) . . . arg ( j ωτ pN + 1 ) arg ( H ( ) ) = arg ( K ) + arg ( j ωτ z1 + 1 ) + arg ( j ωτ z2 + 1 ) + . . . + arg ( j ωτ zM + 1 ) arg ( j ωτ p1 + 1 ) arg ( j ωτ p2 + 1 ) . . . arg ( j ωτ pN + 1 ) alignl { stack { size 12{"arg" $$H \( jω$$ \) ="arg" $$K$$ +"arg" $$j ital "ωτ" rSub { size 8{z1} } +1$$ +"arg" $$j ital "ωτ" rSub { size 8{z2} } +1$$ + "." "." "." +"arg" $$j ital "ωτ" rSub { size 8{ ital "zM"} } +1$$ } {} # matrix { matrix { matrix { matrix { {} # {} } {} # {} } {} # {} } {} # {} } - "arg" $$j ital "ωτ" rSub { size 8{p1} } +1$$ - "arg" $$j ital "ωτ" rSub { size 8{p2} } +1$$ - "." "." "." - "arg" $$j ital "ωτ" rSub { size 8{ ital "pN"} } +1$$ {} } } {}

4/ Logarithmic magnitude

Taking twenty times the logarithm of the magnitude yields

20 log 10 H ( ) = 20 log 10 K + 20 log 10 j ωτ z1 + 1 + 20 log 10 j ωτ z2 + 1 + . . . + 20 log 10 j ωτ zM + 1 20 log 10 j ωτ p1 + 1 20 log 10 j ωτ p2 + 1 . . . 20 log 10 j ωτ pN + 1 20 log 10 H ( ) = 20 log 10 K + 20 log 10 j ωτ z1 + 1 + 20 log 10 j ωτ z2 + 1 + . . . + 20 log 10 j ωτ zM + 1 20 log 10 j ωτ p1 + 1 20 log 10 j ωτ p2 + 1 . . . 20 log 10 j ωτ pN + 1 alignl { stack { size 12{"20""log" rSub { size 8{"10"} } \lline H $$jω$$ \lline ="20""log" rSub { size 8{"10"} } \lline K \lline +"20""log" rSub { size 8{"10"} } \lline j ital "ωτ" rSub { size 8{z1} } +1 \lline +"20""log" rSub { size 8{"10"} } \lline j ital "ωτ" rSub { size 8{z2} } +1 \lline + "." "." "." +"20""log" rSub { size 8{"10"} } \lline j ital "ωτ" rSub { size 8{ ital "zM"} } +1 \lline } {} # matrix { matrix { matrix { matrix { {} # {} } {} # {} } {} # {} } {} # {} } - "20""log" rSub { size 8{"10"} } \lline j ital "ωτ" rSub { size 8{p1} } +1 \lline - "20""log" rSub { size 8{"10"} } \lline j ital "ωτ" rSub { size 8{p2} } +1 \lline - "." "." "." - "20""log" rSub { size 8{"10"} } \lline j ital "ωτ" rSub { size 8{ ital "pN"} } +1 \lline {} } } {}

Note than both the logarithmic magnitude and the angle are expressed as sums of terms of the form

± 20 log 10 j ωτ + 1 and ± arg ( j ωτ + 1 ) ± 20 log 10 j ωτ + 1 and ± arg ( j ωτ + 1 ) size 12{ +- "20 log" rSub { size 8{"10"} } \lline j ital "ωτ" +1 \lline " " matrix { {} # {} } " and " matrix { {} # {} } +- " arg" $$j ital "ωτ" +1$$ } {}

Therefore, to plot the frequency response we need to add terms of the above form.

5/ Decibels

It is common to plot frequency responses as Bode diagrams whose magnitude is expressed in decibels. The decibel, denoted by dB, is defined as 20 log10Η20 log10Η size 12{"20 log" rSub { size 8{"10"} } \lline Η \lline } {}. The following table gives decibel equivalents for a few quantities.

How many decibels correspond to |H| = 50? Express |H| =100/2. Then

20 log 10 ( 100/2 ) = 20 log 10 100 - 20log 10 2 40 - 6 = 34 dB 20 log 10 ( 100/2 ) = 20 log 10 100 - 20log 10 2 40 - 6 = 34 dB size 12{"20 log" rSub { size 8{"10"} } $$"100/2"$$ =" 20 log" rSub { size 8{"10"} } " 100 - 20log" rSub { size 8{"10"} } " 2" approx " 40" "- 6"" = ""34" "dB"} {}

6/ Asymptotes

To plot the frequency response of a system with real poles and zeros, we need to plot terms of the form

± 20 log 10 1 + j ωτ and ± arg ( 1 + j ωτ ) ± 20 log 10 1 + j ωτ and ± arg ( 1 + j ωτ ) size 12{ +- "20 log" rSub { size 8{"10"} } \lline 1+j ital "ωτ" \lline " " matrix { {} # {} } " and " matrix { {} # {} } " " +- " arg" $$1+j ital "ωτ"$$ } {}

The low and high frequency asymptotes are

7/ Corner frequency

At ω= ωc = 1/τω= ωc = 1/τ size 12{ω "= "ω rSub { size 8{c} } " = 1"/τ} {}, called the corner or cut-off frequency,

• the low- and high-frequency asymptotes intersect,
• the magnitude is

± 20 log 10 1+j ωτ = ± 20 log 10 1+j1 = ± 20 log 10 2 1/2 ± 3 dB ± 20 log 10 1+j ωτ = ± 20 log 10 1+j1 = ± 20 log 10 2 1/2 ± 3 dB size 12{ +- "20 log" rSub { size 8{"10"} } \lline "1+j" ital "ωτ" \lline " = " +- "20 log" rSub { size 8{"10"} } \lline "1+j1" \lline " = " +- "20 log" rSub { size 8{"10"} } 2 rSup { size 8{"1/2"} } " " +- "3 dB"} {}

• the angle is

± arg ( 1+ j ωτ ) = ± arg ( 1+ j ) = ± π / 4 ± arg ( 1+ j ωτ ) = ± arg ( 1+ j ) = ± π / 4 size 12{ +- "arg" $$"1+"j ital "ωτ"$$ " = " +- "arg" $$"1+"j$$ " = " +- π/4} {}

Example — first-order lowpass system

First-order low pass systems arise in a large variety of physical contexts. For example,

For the parameters M = B = R = C = 1, the frequency responses for the two systems are

H ( ) = V ( ) F ( ) = 1 + 1 and H ( ) = V 0 ( ) V i ( ) = 1 + 1 H ( ) = V ( ) F ( ) = 1 + 1 and H ( ) = V 0 ( ) V i ( ) = 1 + 1 size 12{H $$jω$$ = { {V $$jω$$ } over {F $$jω$$ } } = { {1} over {jω+1} } matrix { {} # {} } ital "and" matrix { {} # {} } H $$jω$$ = { {V rSub { size 8{0} } $$jω$$ } over {V rSub { size 8{i} } $$jω$$ } } = { {1} over {jω+1} } } {}

Magnitude

For

H ( ) = 1 + 1 H ( ) = 1 + 1 size 12{H $$jω$$ = { {1} over {jω+1} } } {}

the low-frequency asymptote has a slope of 0 and an intercept of 0 dB and the high-frequency asymptote has a slope of -20 dB/decade and an intercept of 0 dB at the corner frequency.

The corner frequency is 1 rad/sec and the bandwidth is 1 rad/sec. The two asymptotes intersect at ω = 1 where 20 log10Η()= -3dB20 log10Η()= -3dB size 12{"20 log" rSub { size 8{"10"} } \lline Η $$"jω"$$ \lline "= ""-3" ital "dB"} {}

H ( ) = 1 + 1 H ( ) = 1 + 1 size 12{H $$jω$$ = { {1} over {jω+1} } } {}

the low- and high-frequency asymptotes of the angle of the frequency response are 0 and 900900 size 12{-"90" rSup { size 8{0} } } {}(−π/2 radians), respectively. The angle is 450450 size 12{-"45" rSup { size 8{0} } } {} at the corner frequency (1 rad/sec).

A line drawn from the low frequency asymptote a decade below the corner frequency to the high frequency asymptote a decade above the corner frequency approximates the angle of the frequency response.

Physical interpretation

With M = B = 1, the frequency response is

H ( ) = V ( ) F ( ) = 1 + 1 H ( ) = V ( ) F ( ) = 1 + 1 size 12{H $$jω$$ = { {V $$jω$$ } over {F $$jω$$ } } = { {1} over {jω+1} } } {}

At low frequencies, |H(jω)| → 1 and arg H(jω) → 0. The inertia of the mass is negligible, and the damping force dominates so that the external force is proportional to velocity.

At high frequencies, |H(jω)| → 1/ω and arg H(jω) → 900900 size 12{-"90" rSup { size 8{0} } } {}. The inertia of the mass dominates so that the acceleration is proportional to external force and the velocity decreases as frequency increases.

With R = C = 1, the frequency response is

{} H ( ) = V o ( ) V i ( ) = 1 + 1 H ( ) = V o ( ) V i ( ) = 1 + 1 size 12{H $$jω$$ = { {V rSub { size 8{o} } $$jω$$ } over {V rSub { size 8{i} } $$jω$$ } } = { {1} over {jω+1} } } {}

At low frequencies, |H(jω)| → 1 and arg H(jω) → 0. The impedance of the capacitance is large so that all the input voltage appears at the output.

At high frequencies, |H(jω)| → 1/ω and arg H(jω) → 900900 size 12{-"90" rSup { size 8{0} } } {}. The impedance of the capacitance is small so that the current is determined by the resistance and the output voltage is determined by the impedance of the capacitance which decreases as frequency increases.

Two-minute miniquiz problem

Problem 6-1

Given

H ( s ) = s ( s + 1 ) ( s + 10 ) H ( s ) = s ( s + 1 ) ( s + 10 ) size 12{H $$s$$ = { {s} over { $$s+1$$ $$s+"10"$$ } } } {}

Determine the magnitude of the frequency response in the form of a Bode diagram.

Solution

H ( ) = 1 10 ( + 1 ) ( / 10 + 1 ) H ( ) = 1 10 ( + 1 ) ( / 10 + 1 ) size 12{H $$jω$$ = { {1} over {"10"} } left [ { {jω} over { $$jω+1$$ $$jω/"10"+1$$ } } right ]} {}

V. SIGNAL PROCESSING WITH FILTERS

1/ Separation of narrowband signals

The input consists of a sum of two sinusoids, more generally the sum of two narrow-band signals such as the signals transmitted by two radio stations. The objective is to devise filters

to separate these two signals. All waveforms have normalized amplitudes.

2/ Extraction of narrow-band signal from wide-band noise

The input consists of a sinusoid, more generally a narrow-band signal, plus some wide-band noise. The objective is to extract the signal from the noise. All waveforms have normalized amplitudes.

3/ Reduction of narrow-band noise from wide-band signal

The input consists of a wide-band signal, in this case an ecg signal recorded from the surface of the chest, plus some narrowband noise, such as pickup from the power lines. The objective is to remove the narrow-band noise. All waveforms have normalized amplitudes.

VI. LOWPASS AND HIGHPASS FILTERS

1/ First-order lowpass and highpass filters

where H(s) = Y (s)/X(s) and RC = 1.

2/ Use of first-order lpf and hpf for signal separation

How much unwanted signal occurs in each output channel? That depends on the frequency separation between the signals. If the two signals are about a decade apart then the attenuation of the unwanted signal will be at most 20 dB with first-order lowpass and highpass filters.

3/ Higher-order lowpass filters, Butterworth filters

If two signals have a small frequency separation or if attenuation of the unwanted signal needs to be very large, higher-order filters are required. The pole-zero diagrams of the class of lowpass Butterworth filters of order 1-9 are shown below.

The system function for the nth-order Butterworth filter, Hn(s)Hn(s) size 12{H rSub { size 8{n} } $$s$$ } {}, is obtained from the property

H n ( s ) H n ( s ) = 1 1 + s c 2n H n ( s ) H n ( s ) = 1 1 + s c 2n size 12{H rSub { size 8{n} } $$s$$ H rSub { size 8{n} } $$- s$$ = { {1} over {1+ left [ { {s} over {jω rSub { size 8{c} } } } right ] rSup { size 8{2n} } } } } {}

where ωcωc size 12{ω rSub { size 8{c} } } {} is the cut-off frequency.

Each additional order of Butterworth filter adds an additional attenuation of −20 dB/decade. The frequency response for the nth-order Butterworth filter, Hn()Hn() size 12{H rSub { size 8{n} } $$jω$$ } {}, is obtained from the property

( H n ( ) ) 2 = 1 1 + ω ω c 2n ( H n ( ) ) 2 = 1 1 + ω ω c 2n size 12{ $$\lline H rSub { size 8{n} } \( jω$$ \lline \) rSup { size 8{2} } = { {1} over {1+ left [ { {ω} over {ω rSub { size 8{c} } } } right ] rSup { size 8{2n} } } } } {}

The frequency responses are shown for n in the range 1-9 and for ωc= 1ωc= 1 size 12{ω rSub { size 8{c} } "= "1} {}.

VII. RESONANCE AND BANDPASS FILTERS

1/ Resonant systems arise in many physical contexts

M is mass, B is a friction constant, K is a spring constant, f(t) is an external force, and v(t) is the velocity of the mass.

A second-order system function relates the velocity to the force

H ( s ) = V ( s ) F ( s ) = 1 M s s 2 + B M s + K M H ( s ) = V ( s ) F ( s ) = 1 M s s 2 + B M s + K M size 12{H $$s$$ = { {V $$s$$ } over {F $$s$$ } } = { { { {1} over {M} } s} over {s rSup { size 8{2} } + { {B} over {M} } s+ { {K} over {M} } } } } {}

Such a mechanical system yields a damped oscillation in response to a force provided the damping is not too large. For example, a tuning fork may be modeled by such a mechanical system.

2/ RLC filter

Electric networks also show a similar system function. The impedance of the RLC circuit is also of second-order

Z ( s ) = V ( s ) I ( s ) = 1 sC + 1 R + 1 sL = sL s 2 LC + s L R + 1 Z ( s ) = V ( s ) I ( s ) = 1 sC + 1 R + 1 sL = sL s 2 LC + s L R + 1 size 12{Z $$s$$ = { {V $$s$$ } over {I $$s$$ } } = { {1} over { ital "sC"+ { {1} over {R} } + { {1} over { ital "sL"} } } } = { { ital "sL"} over {s rSup { size 8{2} } ital "LC"+s { {L} over {R} } +1} } } {}

3/ Resonance parameterized

We can parameterize these and any second-order systems efficiently. We illustrate with the electrical network.

Z ( s ) = G s ω 0 s ω 0 2 + 1 Q s ω 0 + 1 Z ( s ) = G s ω 0 s ω 0 2 + 1 Q s ω 0 + 1 size 12{Z $$s$$ =G { { left [ { {s} over {ω rSub { size 8{0} } } } right ]} over { left [ { {s} over {ω rSub { size 8{0} } } } right ] rSup { size 8{2} } + { {1} over {Q} } left [ { {s} over {ω rSub { size 8{0} } } } right ]+1} } } {}

Where

ω 0 = 1 ( LC ) 1 / 2 ( Undamped resonant frequency ) Q = R C L 1 / 2 ( Quality of resonance ) G = C L 1 / 2 ( Gain cons tan t ) ω 0 = 1 ( LC ) 1 / 2 ( Undamped resonant frequency ) Q = R C L 1 / 2 ( Quality of resonance ) G = C L 1 / 2 ( Gain cons tan t ) alignl { stack { size 12{ω rSub { size 8{0} } = { {1} over { $$ital "LC"$$ rSup { size 8{1/2} } } } matrix { {} # {} } $$ital "Undamped" matrix { {} # {} } ital "resonant" matrix { {} # {} } ital "frequency"$$ } {} # Q=R left [ { {C} over {L} } right ] rSup { size 8{1/2} } matrix { {} # {} } $$ital "Quality" matrix { {} # {} } ital "of" matrix { {} # {} } ital "resonance"$$ {} # G= left [ { {C} over {L} } right ] rSup { size 8{1/2} } matrix { {} # {} } $$ital "Gain" matrix { {} # {} } ital "cons""tan"t$$ {} } } {}

4/ Normalized resonance function

Z ( s ) = G s ω 0 s ω 0 2 + 1 Q s ω 0 + 1 = GR s ω 0 Z ( s ) = G s ω 0 s ω 0 2 + 1 Q s ω 0 + 1 = GR s ω 0 size 12{Z $$s$$ =G { { left [ { {s} over {ω rSub { size 8{0} } } } right ]} over { left [ { {s} over {ω rSub { size 8{0} } } } right ] rSup { size 8{2} } + { {1} over {Q} } left [ { {s} over {ω rSub { size 8{0} } } } right ]+1} } = ital "GR" left [ { {s} over {ω rSub { size 8{0} } } } right ]} {}

where

R ( s ) = s s 2 + 1 Q s + 1 R ( s ) = s s 2 + 1 Q s + 1 size 12{R $$s$$ = { {s} over {s rSup { size 8{2} } + { {1} over {Q} } s+1} } } {}

Therefore, Z(s) and R(s) are the same except for differences in scale. R(s), which is a universal second-order function, depends on only one parameter, Q.

5/ Properties of R(s)

R(s) has a zero at s = 0 and poles at

{ p 1 = 1 2Q + 1 2Q 2 1 1 / 2 p 2 = 1 2Q 1 2Q 2 1 1 / 2 { p 1 = 1 2Q + 1 2Q 2 1 1 / 2 p 2 = 1 2Q 1 2Q 2 1 1 / 2 size 12{ left lbrace matrix { p rSub { size 8{1} } = - left [ { {1} over {2Q} } right ]+ left [ left [ { {1} over {2Q} } right ] rSup { size 8{2} } - 1 right ] rSup { size 8{1/2} } {} ## p rSub { size 8{2} } = - left [ { {1} over {2Q} } right ] - left [ left [ { {1} over {2Q} } right ] rSup { size 8{2} } - 1 right ] rSup { size 8{1/2} } } right none } {}

As Q is varied the poles trace out a trajectory in the s-plane called a root-locus plot.

• For 0 < Q << 1/2, the poles are near 0, and −1/Q.
• For Q ≤ 1/2, the poles are on the real axis.
• For Q = 1/2, both poles are at −1.
• For Q > 1/2, the poles are complex and the locus is circular.

6/ High-Q filter

Recall that for the parallel RLC circuit

Q = R C L 1 / 2 Q = R C L 1 / 2 size 12{Q=R left [ { {C} over {L} } right ] rSup { size 8{1/2} } } {}

When R is arbitrarily large, Q becomes arbitrarily large and we call the network a high-Q system. Note also that in this limit, the current through the resistance becomes arbitrarily small, and it can be shown that the network dissipates relatively little energy.

We shall examine the frequency response of the network in this high-Q limit.

7/ Low-frequency asymptote for the high-Q system

R(s) has one zero and two complex conjugate poles, i.e.,

R ( s ) = s ( s p ) ( s p ) R ( s ) = s ( s p ) ( s p ) size 12{R $$s$$ = { {s} over { $$s - p$$ $$s - p rSup { size 8{*} }$$ } } } {}

A vector diagram is used to interpret the behavior of the system at low frequencies.

At low frequencies, ω<<1, and the system function evaluated at s = jω is

R ( ) ( j ) ( + j ) = R ( ) ( j ) ( + j ) = size 12{R $$jω$$ approx { {jω} over { $$- j$$ $$+j$$ } } =jω} {}

8/ High-frequency asymptote for the high-Q system

We can use a vector diagram to interpret the behavior of the system at high frequencies as follows.

At high frequencies, ω _ 1, and the system function evaluated at s = jω is

R ( ) ( ) ( ) = 1 R ( ) ( ) ( ) = 1 size 12{R $$jω$$ approx { {jω} over { $$jω$$ $$jω$$ } } = { {1} over {jω} } } {}

9/ Behavior of the high-Q system near the resonant frequency

For ω ≈ 1

R ( ) j1 ( ρ ) ( j2 ) = 1 R ( ) j1 ( ρ ) ( j2 ) = 1 size 12{R $$jω$$ approx { {j1} over { $$ρ$$ $$j2$$ } } = { {1} over {2ρ} } } {}

For w = 1

R ( j1 ) 1 2 ( 1 / ( 2Q ) ) = Q and R ( j1 ) 0 R ( j1 ) 1 2 ( 1 / ( 2Q ) ) = Q and R ( j1 ) 0 size 12{ \lline R $$j1$$ \lline approx { {1} over {2 $$1/ \( 2Q$$ \) } } =Q matrix { {} # {} } ital "and" matrix { {} # {} } ∠R $$j1$$ approx 0} {}

For w=1±12Qw=1±12Q size 12{w=1 +- { {1} over {2Q} } } {}

R j 1 ± 1 2Q 1 2 ( 1 / ( 2Q ) ) 2 1 / 2 = Q 2 1 / 2 R j 1 ± 1 2Q π 4 R j 1 ± 1 2Q 1 2 ( 1 / ( 2Q ) ) 2 1 / 2 = Q 2 1 / 2 R j 1 ± 1 2Q π 4 alignl { stack { size 12{ lline R left [j left [1 +- { {1} over {2Q} } right ] right ] rline approx { {1} over {2 $$1/ \( 2Q$$ \) 2 rSup { size 8{1/2} } } } = { {Q} over {2 rSup { size 8{1/2} } } } } {} # ∠R left [j left [1 +- { {1} over {2Q} } right ] right ] approx -+ { {π} over {4} } {} } } {}

10/ Frequency response of the high-Q system

The frequency response, R(jω), is shown in linear coordinates near the resonant frequency (ω = 1). This bandpass filter passes frequencies maximally in a bandwidth BW near the resonant frequency and attenuates frequencies outside that band.

R(jω) is shown in logarithmic coordinates for Q = 10.

The low-frequency asymptote has a magnitude slope of +20 dB/decade and an angle of +90o and the high-frequency asymptote slope is −20 dB/decade and an angle of −90o

11/ Constant-Q filters

We plot the frequency response for bandpass filters with the same value of Q but different resonant frequencies.

For

Z ( s ) = G s ω 0 s ω 0 2 + 1 Q s ω 0 + 1 Z ( s ) = G s ω 0 s ω 0 2 + 1 Q s ω 0 + 1 size 12{Z $$s$$ =G { { left [ { {s} over {ω rSub { size 8{0} } } } right ]} over { left [ { {s} over {ω rSub { size 8{0} } } } right ] rSup { size 8{2} } + { {1} over {Q} } left [ { {s} over {ω rSub { size 8{0} } } } right ]+1} } } {}

we plot frequency responses for G = 1, Q = 10, ω0=1ω0=1 size 12{ω rSub { size 8{0} } =1} {}, 101/2101/2 size 12{"10" rSup { size 8{1/2} } } {}, 10, and 103/2103/2 size 12{"10" rSup { size 8{3/2} } } {},

12/ Constant- ωo filters

We plot the frequency response for bandpass filters with the same value of ωo but different values of Q.

For

Z ( s ) = G s ω 0 s ω 0 2 + 1 Q s ω 0 + 1 Z ( s ) = G s ω 0 s ω 0 2 + 1 Q s ω 0 + 1 size 12{Z $$s$$ =G { { left [ { {s} over {ω rSub { size 8{0} } } } right ]} over { left [ { {s} over {ω rSub { size 8{0} } } } right ] rSup { size 8{2} } + { {1} over {Q} } left [ { {s} over {ω rSub { size 8{0} } } } right ]+1} } } {}

we plot frequency responses for G = 1/Q, ω = 1. The parameter is Q = 1, 10, and 100.

Two-minute miniquiz problem

Problem 7-1 Frequency response of a resonant system

Given

H ( s ) = 10 s ( s + 1 ) 2 + 100 H ( s ) = 10 s ( s + 1 ) 2 + 100 size 12{H $$s$$ ="10" { {s} over { $$s+1$$ rSup { size 8{2} } +"100"} } } {}

Determine (approximately)

a) the resonant frequency,

b) the bandwidth,

c) |H(jω)| at the resonant frequency.

Solution

The system function has a zero at s = 0 and poles at s = −1 ± j10. Near resonance the vector diagram looks as follows

a) The resonant frequency is ω010rad/sω010rad/s size 12{ω rSub { size 8{0} } approx "10" ital "rad"/s} {}.

b) From a vector interpretation, the frequency response has 3 dB points at 9 and 11 rad/s. Hence, the bandwidth is 2 rad/s.

c) From a vector interpretation,

H ( j 10 ) 10 j 10 ( 1 ) ( j 20 ) = 5 H ( j 10 ) 10 j 10 ( 1 ) ( j 20 ) = 5 size 12{H $$j"10"$$ approx "10" { {j"10"} over { $$1$$ $$j"20"$$ } } =5} {}

13/ Use of bpf for extraction of narrow-band signals in wideband noise

A resonant circuit acting as a narrow-band or bandpass filter (BPF) can be used to extract a narrow-band signal from wideband noise. The resonant frequency of the filter is set to equal

the center frequency of the signal. The value of Q is chosen to determine the desired amount of attenuation of the wide-band noise.

VIII. SECOND-ORDER NOTCH FILTERS

1/ Notch filter — system function and pole-zero diagram

A second-order notch filter has the form shown below.

H ( s ) = s ω 0 2 + 1 s ω 0 2 + 1 Q s ω 0 2 + 1 H ( s ) = s ω 0 2 + 1 s ω 0 2 + 1 Q s ω 0 2 + 1 size 12{H $$s$$ = { { left [ { {s} over {ω rSub { size 8{0} } } } right ] rSup { size 8{2} } +1} over { left [ { {s} over {ω rSub { size 8{0} } } } right ] rSup { size 8{2} } + { {1} over {Q} } left [ { {s} over {ω rSub { size 8{0} } } } right ] rSup { size 8{2} } +1} } } {}

2/ Notch filter — frequency response

H ( ) = 1 ω ω 0 2 1 ω ω 0 2 + j 1 Q ω ω 0 H ( ) = 1 ω ω 0 2 1 ω ω 0 2 + j 1 Q ω ω 0 size 12{H $$jω$$ = { {1 - left [ { {ω} over {ω rSub { size 8{0} } } } right ] rSup { size 8{2} } } over {1 - left [ { {ω} over {ω rSub { size 8{0} } } } right ] rSup { size 8{2} } +j { {1} over {Q} } left [ { {ω} over {ω rSub { size 8{0} } } } right ]} } } {}

The notch filter has a frequency response

H ( ) = { 0 for ω = ω 0 1 for ω << ω 0 and for ω >> ω 0 H ( ) = { 0 for ω = ω 0 1 for ω << ω 0 and for ω >> ω 0 size 12{H $$jω$$ = left lbrace matrix { 0 matrix { matrix { {} # {} } {} # {} } matrix { {} # {} } ital "for" matrix { {} # {} } ω=ω rSub { size 8{0} } {} ## 1 matrix { {} # {} } ital "for" matrix { {} # {} } ω"<<"ω rSub { size 8{0} } matrix { {} # {} } ital "and" matrix { {} # {} } ital "for" matrix { {} # {} } ω">>"ω rSub { size 8{0} } {} } right none } {}

3/ Use of notch filter to attenuate narrow-band noise in a wide-band signal

A notch filter (NF) can be used to attenuate a narrow-band noise in a wide-band signal. The resonant frequency of the filter is set to equal the center frequency of the noise. The value of Q is chosen to determine the desired amount of attenuation of the narrow-band noise. In the example shown, a LPF and HPF can also be used if there is sufficient frequency separation between desirable and undesirable components of the signals.

IX. CONCLUSIONS

The frequency response

• is the system function evaluated along the jω-axis,
• has a simple geometric interpretation in the s-plane,
• describes the filtering of an input sinusoid to an LTI system,
• has asymptotes that are easily sketched in a Bode diagram when the poles and zeros lie on the negative real axis.

Filters have been illustrated to perform a variety of important signal processing tasks. The filters we have introduced are

• lowpass and highpass filters both first-order and higher order,
• bandpass filters consisting of resonant systems,
• notch filters.

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