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# FREQUENCY RESPONSE OF LTI (LSI) SYSTEMS

Module by: Nguyen Huu Phuong. E-mail the author

Up to now the discussion has been on discrete-time signals. As a matter of fact, most the discussion so far also applies to systems (assumed to be LTI or LSI). However there are some differences, e.g. the meaning of time convolution.

A system is characterized by its impulse h(n)h(n) size 12{h $$n$$ } {} whose DTFT transform is

H(ω)=n=h(n)ejωnH(ω)=n=h(n)ejωn size 12{H $$ω$$ = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h $$n$$ e rSup { size 8{ - jωn} } } } {}
(1)

And the inverse DTFT is

hn=1ππHωejωnhn=1ππHωejωn size 12{h left (n right )= { {1} over {2π} } Int rSub { size 8{ - π} } rSup { size 8{π} } {H left (ω right )e rSup { size 8{jωn} } dω} } {}
(2)

H(ω)H(ω) size 12{H $$ω$$ } {} is called the frequency response or frequency characteristic of the system. It is the frequency characterization of the system whereas the impulse response is the time characterization.

## Frequency response

Now we use the time convolution property (or convolution theorem) to map the output y(n)y(n) size 12{y $$n$$ } {} in time domain to its transform Y(ω)Y(ω) size 12{Y $$ω$$ } {} in the frequency domain (Fig.3.28) :

y(n)=x(n)h(n)Y(ω)=X(ω)H(ω)y(n)=x(n)h(n)Y(ω)=X(ω)H(ω) size 12{y $$n$$ =x $$n$$ *h $$n$$ ↔Y $$ω$$ =X $$ω$$ H $$ω$$ } {}
(3)

Or

H(ω)=Y(ω)X(ω)H(ω)=Y(ω)X(ω) size 12{H $$ω$$ = { {Y $$ω$$ } over {X $$ω$$ } } } {}
(4)

The frequency response H(ω)H(ω) size 12{H $$ω$$ } {}is usually a complex quantily, so we write

H(ω)=HR(ω)+jHI(ω)=H(ω)eωH(ω)=HR(ω)+jHI(ω)=H(ω)eω size 12{H $$ω$$ =H rSub { size 8{R} } $$ω$$ + ital "jH" rSub { size 8{I} } $$ω$$ = lline H $$ω$$ rline e rSup { size 8{jΦ left (ω right )} } } {}
(5)

where

H(ω)=HR2(ω)+HI2ωH(ω)=HR2(ω)+HI2ω size 12{ lline H $$ω$$ rline = sqrt {H rSub { size 8{R} } rSup { size 8{2} } $$ω$$ +H rSub { size 8{I} } rSup { size 8{2} } left (ω right )} } {}
(6)

and

Φ(ω)=tan1HI(ω)HR(ω)Φ(ω)=tan1HI(ω)HR(ω) size 12{Φ $$ω$$ ="tan" rSup { size 8{ - 1} } { {H rSub { size 8{I} } $$ω$$ } over {H rSub { size 8{R} } $$ω$$ } } } {}
(7)

H(ω)H(ω) size 12{ lline H $$ω$$ rline } {} is the magnitude response anf Φ(ω)Φ(ω) size 12{Φ $$ω$$ } {} is the phase response. If the impulse response h(n)h(n) size 12{h $$n$$ } {} is real-valued then

H(ω)=H(ω)andΦ(ω)=Φ(ω)H(ω)=H(ω)andΦ(ω)=Φ(ω) size 12{ lline H $$- ω$$ rline = lline H $$ω$$ rline matrix { {} # {} } ital "and" matrix { {} # {} } Φ $$- ω$$ = - Φ $$ω$$ } {}
(8)

The frequency response of a system exists if the system is BIBO stable, i.e. (Equation (2.15))

n=hn<n=hn< size 12{ Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { lline h left (n right ) rline } < infinity } {}
(9)

Example 3.7.1

Find the frequency response of a system whose input-output difference equation is

y ( n ) = A [ x ( n 2 ) + x ( n 1 ) + x ( n ) + x ( n + 1 ) + x ( n + 2 ) ] y ( n ) = A [ x ( n 2 ) + x ( n 1 ) + x ( n ) + x ( n + 1 ) + x ( n + 2 ) ] size 12{y $$n$$ =A $x $$n - 2$$ +x $$n - 1$$ +x $$n$$ +x $$n+1$$ +x $$n+2$$$ } {}

where A is a constant.

Solution

First the impulse response h(n)h(n) size 12{h $$n$$ } {} is just the output y(n)y(n) size 12{y $$n$$ } {} when the input is x(n)=δ(n)x(n)=δ(n) size 12{x $$n$$ =δ $$n$$ } {}, thus

h ( n ) = A [ δ ( n 2 ) + δ ( n 1 ) + δ ( n ) + δ ( n + 1 ) + δ ( n + 2 ) ] h ( n ) = A [ δ ( n 2 ) + δ ( n 1 ) + δ ( n ) + δ ( n + 1 ) + δ ( n + 2 ) ] size 12{h $$n$$ =A $δ $$n - 2$$ +δ $$n - 1$$ +δ $$n$$ +δ $$n+1$$ +δ $$n+2$$$ } {}

It turns out that this impulse response is the same as the signal in Example 3.4.1, hence the frequency response of the system is

H ( ω ) = n = 2 2 h ( n ) e jωn = A ( 1 + 2 cos ω + 2 cos ) H ( ω ) = n = 2 2 h ( n ) e jωn = A ( 1 + 2 cos ω + 2 cos ) size 12{H $$ω$$ = Sum cSub { size 8{n= - 2} } cSup { size 8{2} } {h $$n$$ e rSup { size 8{ - jωn} } } =A $$1+2"cos"ω+2"cos"2ω$$ } {}

The system is a low-pass filter.

Example 3.7.2

A system has impulse response.

h ( n ) = 0 . 8 n u ( n ) h ( n ) = 0 . 8 n u ( n ) size 12{h $$n$$ =0 "." 8 rSup { size 8{n} } u $$n$$ } {}

Plot the frequency responses HR(ω),HI(ω),H(ω)HR(ω),HI(ω),H(ω) size 12{H rSub { size 8{R} } $$ω$$ ,H rSub { size 8{I} } $$ω$$ , lline H $$ω$$ rline } {} and Φ(ω)Φ(ω) size 12{Φ $$ω$$ } {}.

Solution

This problem is the same as Example 3.4.2. The frequency response is

H ( ω ) = n = h ( n ) e jn ω = n = 0 ( 0,8 e ) n = 1 1 0 . 8 e H ( ω ) = n = h ( n ) e jn ω = n = 0 ( 0,8 e ) n = 1 1 0 . 8 e size 12{H $$ω$$ = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h $$n$$ e rSup { size 8{ - ital "jn"ω} } = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { $$0,8e rSup { size 8{ - jω} }$$ rSup { size 8{n} } } } = { {1} over {1 - 0 "." 8e rSup { size 8{ - jω} } } } } {}

In order to compute the real and imaginary frequency responses we write

H ( ω ) = 1 0 . 8 e ( 1 0 . 8 e ) ( 1 0 . 8 e ) = 1 0 . 8 cos ω j0 . 8 sin ω 1 . 64 1 . 6 cos ω H ( ω ) = 1 0 . 8 e ( 1 0 . 8 e ) ( 1 0 . 8 e ) = 1 0 . 8 cos ω j0 . 8 sin ω 1 . 64 1 . 6 cos ω size 12{H $$ω$$ = { {1 - 0 "." 8e rSup { size 8{jω} } } over { $$1 - 0 "." 8e rSup { size 8{ - jω} }$$ $$1 - 0 "." 8e rSup { size 8{jω} }$$ } } = { {1 - 0 "." 8"cos"ω - j0 "." 8"sin"ω} over {1 "." "64" - 1 "." 6"cos"ω} } } {}

From this,

H R ( ω ) = 1 0 . 8 cos ω 1 . 64 1 . 6 cos ω H R ( ω ) = 1 0 . 8 cos ω 1 . 64 1 . 6 cos ω size 12{H rSub { size 8{R} } $$ω$$ = { {1 - 0 "." 8"cos"ω} over {1 "." "64" - 1 "." 6"cos"ω} } } {}

H I ( ω ) = 0 . 8 sin ω 1 . 64 1 . 6 cos ω H I ( ω ) = 0 . 8 sin ω 1 . 64 1 . 6 cos ω size 12{H rSub { size 8{I} } $$ω$$ = { {0 "." 8"sin"ω} over {1 "." "64" - 1 "." 6"cos"ω} } } {}

For the magnitude response H(ω)H(ω) size 12{ lline H $$ω$$ rline } {} we’d better not go from these two components, but rather from the original expression of H(ω)H(ω) size 12{H $$ω$$ } {}:

H ( ω ) = 1 ( 1 0 . 8 cos ω ) + j0 . 8 sin ω H ( ω ) = 1 ( 1 0 . 8 cos ω ) + j0 . 8 sin ω size 12{H $$ω$$ = { {1} over { $$1 - 0 "." 8"cos"ω$$ +j0 "." 8"sin"ω} } } {}

then

H ( ω ) = 1 [ ( 1 0 . 8 cos ω ) 2 + ( 0 . 8 sin ω ) 2 ] 1 2 = 1 [ 1 . 64 1 . 60 cos ω ] 1 2 H ( ω ) = 1 [ ( 1 0 . 8 cos ω ) 2 + ( 0 . 8 sin ω ) 2 ] 1 2 = 1 [ 1 . 64 1 . 60 cos ω ] 1 2 size 12{ lline H $$ω$$ rline = { {1} over { $$$1 - 0 "." 8"cos"ω$$ rSup { size 8{2} } + $$0 "." 8"sin"ω$$ rSup { size 8{2} }$ rSup { size 8{ { {1} over {2} } } } } } = { {1} over { $1 "." "64" - 1 "." "60""cos"ω$ rSup { size 8{ { {1} over {2} } } } } } } {}

The phase response is

Φ ( ω ) = tan 1 0 . 8 sin ω 1 0 . 8 cos ω Φ ( ω ) = tan 1 0 . 8 sin ω 1 0 . 8 cos ω size 12{Φ $$ω$$ = - "tan" rSup { size 8{ - 1} } { {0 "." 8"sin"ω} over {1 - 0 "." 8"cos"ω} } } {}

Fig.3.29 presents all the required spectra.

## Magnitude frequency response on decibel scale

So far the linear scale has been used on the vertical axis (ordinates) to express the of frequency response magnitude. We know in electronics that the logarithmic scale, mainly the decibel (dB) scale, is often used for the horizontal axis (abscissa) to reduce large variations such as from 10 to 109109 size 12{"10" rSup { size 8{9} } } {}. But in DSP (DTSP) the logarithmic scale is used for ordinates to enlarge small variations in amplitudes to make the sidelobes more pronounced.

The magnitude in dBs H(ω)dBH(ω)dB size 12{H $$ω$$ \lline rSub { size 8{ ital "dB"} } } {} is related to the magnitude H(ω)H(ω) size 12{ lline H $$ω$$ rline } {} in linear scale as

H(ω)dB=20log10H(ω)H(ω)dB=20log10H(ω) size 12{ lline H $$ω$$ rline rSub { size 8{ ital "dB"} } ="20""log" rSub { size 8{"10"} } lline H $$ω$$ rline } {}
(10)

See Fig.3.30. Remember when H(ω)=1H(ω)=1 size 12{ lline H $$ω$$ rline =1} {} then H(ω)dB=0H(ω)dB=0 size 12{ lline H $$ω$$ rline rSub { size 8{ ital "dB"} } =0} {}, when H(ω)>1H(ω)>1 size 12{ lline H $$ω$$ rline >1} {} the dBs are positive, when H(ω)<1H(ω)<1 size 12{ lline H $$ω$$ rline <1} {} the dBs are negative. Some examples are as follows.

Observing the logarith variation we see that when H(ω)H(ω) size 12{ lline H $$ω$$ rline } {} is in the range 0 to 0.1, the dB varies extremely fast from size 12{ - infinity } {} to – 20 dB. And when H(ω)H(ω) size 12{ lline H $$ω$$ rline } {} has values from 0.5 upwards the dB slows down. Fig.3.30 materializes this observation. The small variation of H(ω)H(ω) size 12{ lline H $$ω$$ rline } {}around 1 dissappears on H(ω)dBH(ω)dB size 12{ lline H $$ω$$ rline rSub { size 8{ ital "dB"} } } {}, whereas the unseen variation of H(ω)H(ω) size 12{ lline H $$ω$$ rline } {} around 0 is greatly magnified on H(ω)dBH(ω)dB size 12{ lline H $$ω$$ rline rSub { size 8{ ital "dB"} } } {}.

## Eigen-function and eigen-value in DSP systems

Here, the idea is find a signal which preserves its time identity when going through a system. Let’s start with a discrete cosine

x ( n ) = cos ωn x ( n ) = cos ωn size 12{x $$n$$ ="cos"ωn} {}

The corresponding signal out of a system represented by the inpulse response h(n) is

y ( n ) = k = h ( k ) cos ω ( n k ) = k = h ( k ) cos ωk cos ωn + k = h ( k ) sin ωk sin ωn y ( n ) = k = h ( k ) cos ω ( n k ) = k = h ( k ) cos ωk cos ωn + k = h ( k ) sin ωk sin ωn alignl { stack { size 12{y $$n$$ = Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {h $$k$$ "cos"ω $$n - k$$ } } {} # matrix { {} # {} } = left [ Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {h $$k$$ "cos"ωk} right ]"cos"ωn+ left [ Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {h $$k$$ "sin"ωk} right ]"sin"ωn {} } } {}

Both factors in brackets are independent of time as expected but there is an unwanted accompanied sine term. Now let’s test with a complex exponential

x(n)=ejωnx(n)=ejωn size 12{x $$n$$ =e rSup { size 8{jωn} } } {}
(11)

The output is

y ( n ) = k = h ( k ) e n k = k = h ( k ) e jωk e jωn y ( n ) = k = h ( k ) e n k = k = h ( k ) e jωk e jωn size 12{y $$n$$ = Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {h $$k$$ e rSup { size 8{jω left (n - k right )} } } = left [ Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {h $$k$$ e rSup { size 8{ - jωk} } } right ]e rSup { size 8{jωn} } } {}

So the input exponential appears wholly at the output, its time variation does not change. The factor in brackets is just the frequency response H(ω)H(ω) size 12{H $$ω$$ } {}, so

y(n)=H(ω)ejωny(n)=H(ω)ejωn size 12{y $$n$$ =H $$ω$$ e rSup { size 8{jωn} } } {}
(12)

In the mathematical language H(ω)H(ω) size 12{H $$ω$$ } {}is the eigen-value and ejωnejωn size 12{e rSup { size 8{jωn} } } {} the eigen-function. Actually, the phase of the input signal ejωnejωn size 12{e rSup { size 8{jωn} } } {} has been change. For this we write

y ( n ) = H ( ω ) e ω e jωn = H ( ω ) e j ωn + Φ ω y ( n ) = H ( ω ) e ω e jωn = H ( ω ) e j ωn + Φ ω size 12{y $$n$$ = lline H $$ω$$ rline e rSup { size 8{jΦ left (ω right )} } e rSup { size 8{jωn} } = lline H $$ω$$ rline e rSup { size 8{j left (ωn+Φ left (ω right ) right )} } } {}

Example 3.7.3

A filter has impulse response

h ( n ) = 0 . 8 n u ( n ) h ( n ) = 0 . 8 n u ( n ) size 12{h $$n$$ =0 "." 8 rSup { size 8{n} } u $$n$$ } {}

Find the output for the input

1. x ( n ) = 1 . 64 e jn π / 2 x ( n ) = 1 . 64 e jn π / 2 size 12{x $$n$$ =1 "." "64"e rSup { size 8{ ital "jn" {π} slash {2} } } } {}
2. x ( n ) = 2 cos n π / 2 x ( n ) = 2 cos n π / 2 size 12{x $$n$$ =2"cos"n {π} slash {2} } {}

Solution

First we find the filter frequency response

H ( ω ) = n = h ( n ) e jωn = n = 0 ( 0 . 8 e ) n = 1 1 0 . 8 e H ( ω ) = n = h ( n ) e jωn = n = 0 ( 0 . 8 e ) n = 1 1 0 . 8 e size 12{H $$ω$$ = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h $$n$$ e rSup { size 8{ - jωn} } } = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { $$0 "." 8e rSup { size 8{ - jω} }$$ rSup { size 8{n} } } = { {1} over {1 - 0 "." 8e rSup { size 8{ - jω} } } } } {}

Notice the both signals in (a) and (b) have the same angular frequency of ω=π/2ω=π/2 size 12{ω= {π} slash {2} } {}. The frequency response at this frequency is

H π 2 = 1 1 + j0 . 8 = 1 1 0 . 8 e / 2 = 1 1 + j0 . 8 = 1 1 . 64 e j 38 . 66 0 H π 2 = 1 1 + j0 . 8 = 1 1 0 . 8 e / 2 = 1 1 + j0 . 8 = 1 1 . 64 e j 38 . 66 0 size 12{H left ( { {π} over {2} } right )= { {1} over {1+j0 "." 8} } = { {1} over {1 - 0 "." 8e rSup { size 8{ - jπ/2} } } } = { {1} over {1+j0 "." 8} } = { {1} over { sqrt {1 "." "64"} } } e rSup { size 8{ - j"38" "." "66" rSup { size 6{0} } } } } {}

(a) The output signal with respect to this input is

y ( n ) = H π 2 e jn π / 2 = 2 . 5 2 5 e j 26 . 6 0 e jn π / 2 = 5 e j / 2 26 , 6 0 < n < y ( n ) = H π 2 e jn π / 2 = 2 . 5 2 5 e j 26 . 6 0 e jn π / 2 = 5 e j / 2 26 , 6 0 < n < alignl { stack { size 12{y $$n$$ =H left ( { {π} over {2} } right )e rSup { size 8{ ital "jn"π/2} } =2 "." 5 left ( { {2} over { sqrt {5} } } e rSup { size 8{ - j"26" "." 6 rSup { size 6{0} } } } right )e rSup { ital "jn"π/2} } {} # size 12{ matrix { matrix { {} # {} # {} # {} } {} # {} # {} } = sqrt {5} e rSup { size 8{j left (nπ/2 - "26",6 rSup { size 6{0} } right )} } matrix { {} # {} # {} } - infinity <n< infinity } {} } } {}

y ( n ) = H π 2 x ( n ) = 1 . 64 1 . 64 e j ( / 2 38 . 66 0 ) = 1 . 64 e j ( / 2 38 . 66 0 ) y ( n ) = H π 2 x ( n ) = 1 . 64 1 . 64 e j ( / 2 38 . 66 0 ) = 1 . 64 e j ( / 2 38 . 66 0 ) size 12{y $$n$$ =H left ( { {π} over {2} } right )x $$n$$ = { {1 "." "64"} over { sqrt {1 "." "64"} } } e rSup { size 8{j $$nπ/2 - "38" "." "66" rSup { size 6{0} }$$ } } = sqrt {1 "." "64"} e rSup {j $$nπ/2} size 12{ - "38" "." "66" rSup {0} } size 12{$$ }} {}

(b) For the cosinusoidal input, first we write

2 cos n π 2 = e jn π 2 + e jn π 2 2 cos n π 2 = e jn π 2 + e jn π 2 size 12{2"cos"n { {π} over {2} } =e rSup { size 8{ ital "jn" { {π} over {2} } } } +e rSup { size 8{ - ital "jn" { {π} over {2} } } } } {}

Thus the output is

y ( n ) = H π 2 ejn π / 2 + H π 2 e jn π / 2 = 1 1 . 64 e j 38 . 66 0 e jn π / 2 + 1 1 . 64 e j 38 . 60 0 e jn π / 2 = 1 1 . 64 e j ( 38 . 60 0 / 2 ) + e j ( 38 . 66 0 / 2 = 2 1 . 64 cos ( / 2 38 . 66 0 ) y ( n ) = H π 2 ejn π / 2 + H π 2 e jn π / 2 = 1 1 . 64 e j 38 . 66 0 e jn π / 2 + 1 1 . 64 e j 38 . 60 0 e jn π / 2 = 1 1 . 64 e j ( 38 . 60 0 / 2 ) + e j ( 38 . 66 0 / 2 = 2 1 . 64 cos ( / 2 38 . 66 0 ) alignl { stack { size 12{y $$n$$ =H left ( { {π} over {2} } right ) ital "ejn"π/2+H left ( - { {π} over {2} } right )e rSup { size 8{ - ital "jn"π/2} } } {} # matrix { {} # {} # {} } = { {1} over { sqrt {1 "." "64"} } } e rSup { size 8{ - j"38" "." "66" rSup { size 6{0} } } } e rSup { ital "jn"π/2} size 12{+ { {1} over { sqrt {1 "." "64"} } } e rSup {j"38" "." "60" rSup { size 6{0} } } } size 12{e rSup { - ital "jn"π/2} } {} # size 12{ matrix { {} # {} # {} } = { {1} over { sqrt {1 "." "64"} } } left [e rSup {j $$"38" "." "60" rSup { size 6{0} } - nπ/2$$ } size 12{+e rSup { - j $$"38" "." "66" rSup { size 6{0} } - nπ/2} } right ]} {} # size 12{ matrix { {} # {} # {} } = { {2} over { sqrt {1 "." "64"} } } "cos" \( nπ/2 - "38" "." "66" rSup {0} size 12{$$ }} {} } } {}

## Frequency response of systems in cascade or in parallel

In various situations filters are connected in cascade or in parallel, and we would like to know the overall frequency response. By using the associativity and the distributivity of impulse responses, and the convolution theorem of DTFT we can obtain (Fig.3.34):

1. Systems in cascade

H(ω)=H1(ω)H2(ω)...H(ω)=H1(ω)H2(ω)... size 12{H $$ω$$ =H rSub { size 8{1} } $$ω$$ H rSub { size 8{2} } $$ω$$ "." "." "." } {}
(13)

1. Systems in parallel

H(ω)=H1(ω)+H2(ω)+...H(ω)=H1(ω)+H2(ω)+... size 12{H $$ω$$ =H rSub { size 8{1} } $$ω$$ +H rSub { size 8{2} } $$ω$$ + "." "." "." } {}
(14)

## Frequency response in terms of filter coefficients

From the difference equation of general linear filter (Equation 2.21)

y ( n ) = k = 1 N a k y ( n k ) + K = M M b k x ( n k ) y ( n ) = k = 1 N a k y ( n k ) + K = M M b k x ( n k ) size 12{y $$n$$ = Sum cSub { size 8{k=1} } cSup { size 8{N} } {a rSub { size 8{k} } y $$n - k$$ } + Sum cSub { size 8{K= - M} } cSup { size 8{M} } {b rSub { size 8{k} } x $$n - k$$ } } {}

For input ejωnejωn size 12{e rSup { size 8{jωn} } } {}the output is (Equation (3.70))

y ( n ) = H ( ω ) e jωn y ( n ) = H ( ω ) e jωn size 12{y $$n$$ =H $$ω$$ e rSup { size 8{jωn} } } {}

We replace this into the difference equation:

H ( ω ) e jn ω = k = 1 N a k H ( ω ) e n k + k = M M a k e n k H ( ω ) e jn ω = k = 1 N a k H ( ω ) e n k + k = M M a k e n k size 12{H $$ω$$ e rSup { size 8{ ital "jn"ω} } = Sum cSub { size 8{k=1} } cSup { size 8{N} } {a rSub { size 8{k} } H $$ω$$ e rSup { size 8{jω left (n - k right )} } +{}} Sum cSub { size 8{k= - M} } cSup { size 8{M} } {a rSub { size 8{k} } e rSup { size 8{jω left (n - k right )} } } } {}

Thus

H(ω)=k=MMbkejωk1k=1NakejωkH(ω)=k=MMbkejωk1k=1Nakejωk size 12{H $$ω$$ = { { Sum cSub { size 8{k= - M} } cSup { size 8{M} } {b rSub { size 8{k} } e rSup { size 8{ - jωk} } } } over {1 - Sum cSub { size 8{k=1} } cSup { size 8{N} } {a rSub { size 8{k} } e rSup { size 8{ - jωk} } } } } } {}
(15)

This result means the when we know the coefficients of a filter we can write the expression of its frequency response immediately. Conversely, if we know the expression of the frequency response of a system we can write its difference equation.

The normal way to compute the frequency response in to express it as a rational function

H(ω)=N(ω)D(ω)=N(ω)eNωD(ω)eDωH(ω)=N(ω)D(ω)=N(ω)eNωD(ω)eDω size 12{H $$ω$$ = { {N $$ω$$ } over {D $$ω$$ } } = { { lline N $$ω$$ rline e rSup { size 8{jΦ rSub { size 6{N} } left (ω right )} } } over { lline D $$ω$$ rline e rSup {jΦ rSub { size 6{D} } left (ω right )} } } } {}
(16)

Where ΦN(ω)ΦN(ω) size 12{Φ rSub { size 8{N} } $$ω$$ } {} or N(ω)N(ω) size 12{∠N $$ω$$ } {} is the phase of N(ω)N(ω) size 12{N $$ω$$ } {}, and ΦD(N)ΦD(N) size 12{Φ rSub { size 8{D} } $$N$$ } {}or D(ω)D(ω) size 12{∠D $$ω$$ } {} is the phase of D(ω)D(ω) size 12{D $$ω$$ } {}. The magnitude and phase responses are then given respectively by

H(ω)=N(ω)D(ω)H(ω)=N(ω)D(ω) size 12{ lline H $$ω$$ rline = { { lline N $$ω$$ rline } over { lline D $$ω$$ rline } } } {}
(17)

Φ(ω)=ΦN(ω)ΦD(ω)orH(ω)=∠N(ω)D(ω)Φ(ω)=ΦN(ω)ΦD(ω)orH(ω)=∠N(ω)D(ω) size 12{Φ $$ω$$ =Φ rSub { size 8{N} } $$ω$$ - Φ rSub { size 8{D} } $$ω$$ matrix { {} # {} # {} } ital "or" matrix { {} # {} } ∠H $$ω$$ "=∠"N $$ω$$ - ∠D $$ω$$ } {}
(18)

Unless for simple filters (i.e. filters with a small number of coefficients) we must write a computer programme, or use a software such as Matlab, to compute the frequency response and especially the phase response.

Example 3.7.4

An IIR filter has these nonzero coefficients

a 0 = 0 . 04 , a 2 = 0 . 05 , a 4 = 0 . 06 , a 6 = 0 . 11 , a 8 = 0 . 32 , a 9 = 0 . 5, a 10 = 0 . 32 , a 12 = 0 . 11 , a 14 = 0 . 06 , a 16 = 0 . 05 , a 18 = 0 . 04 a 0 = 0 . 04 , a 2 = 0 . 05 , a 4 = 0 . 06 , a 6 = 0 . 11 , a 8 = 0 . 32 , a 9 = 0 . 5, a 10 = 0 . 32 , a 12 = 0 . 11 , a 14 = 0 . 06 , a 16 = 0 . 05 , a 18 = 0 . 04 alignl { stack { size 12{a rSub { size 8{0} } =0 "." "04", matrix { } a rSub { size 8{2} } = - 0 "." "05", matrix { } a rSub { size 8{4} } =0 "." "06", matrix { } a rSub { size 8{6} } = - 0 "." "11", matrix { } a rSub { size 8{8} } =0 "." "32", matrix { } a rSub { size 8{9} } = - 0 "." 5, matrix { } a rSub { size 8{"10"} } =0 "." "32",} {} # a rSub { size 8{"12"} } = - 0 "." "11", matrix { {} } a rSub { size 8{"14"} } =0 "." "06", matrix { {} } a rSub { size 8{"16"} } = - 0 "." "05", matrix { {} } a rSub { size 8{"18"} } =0 "." "04" {} } } {}

Write the difference equation of the filter and find the frequency response.

Solution

The difference equation the FIR filter is

y ( n ) = 0 . 04 x ( n ) 0 . 05 x ( n 2 ) + 0 . 06 x ( n 4 ) 0 . 11 x ( n 6 ) + 0 . 32 x ( n 8 ) 0 . 5x ( n 9 ) + 0 . 32 x ( n 10 ) 0 . 11 x ( n 12 ) + 0 . 06 x ( n 14 ) 0 . 05 x ( n 16 ) + 0 . 04 x ( n 18 ) y ( n ) = 0 . 04 x ( n ) 0 . 05 x ( n 2 ) + 0 . 06 x ( n 4 ) 0 . 11 x ( n 6 ) + 0 . 32 x ( n 8 ) 0 . 5x ( n 9 ) + 0 . 32 x ( n 10 ) 0 . 11 x ( n 12 ) + 0 . 06 x ( n 14 ) 0 . 05 x ( n 16 ) + 0 . 04 x ( n 18 ) alignl { stack { size 12{y $$n$$ =0 "." "04"x $$n$$ - 0 "." "05"x $$n - 2$$ +0 "." "06"x $$n - 4$$ - 0 "." "11"x $$n - 6$$ +0 "." "32"x $$n - 8$$ } {} # size 12{ matrix { {} # {} # {} } - 0 "." 5x $$n - 9$$ +0 "." "32"x $$n - "10"$$ - 0 "." "11"x $$n - "12"$$ +0 "." "06"x $$n - "14"$$ } {} # size 12{ matrix { {} # {} # {} } - 0 "." "05"x $$n - "16"$$ +0 "." "04"x $$n - "18"$$ } {} } } {}

Because all the coefficients are real hence we can quickly find the expressions for the real and imaginary parts:

H R ( ω ) = 0 . 04 cos 0 . 05 cos + 0 . 06 cos 0 . 11 cos + 0 . 32 cos 0 . 5 cos + 0 . 32 cos 10 ω 0 . 11 cos 12 ω + 0 . 06 cos 14 ω 0 . 05 cos 16 ω + 0 . 04 cos 18 ω H R ( ω ) = 0 . 04 cos 0 . 05 cos + 0 . 06 cos 0 . 11 cos + 0 . 32 cos 0 . 5 cos + 0 . 32 cos 10 ω 0 . 11 cos 12 ω + 0 . 06 cos 14 ω 0 . 05 cos 16 ω + 0 . 04 cos 18 ω alignl { stack { size 12{H rSub { size 8{R} } $$ω$$ =0 "." "04""cos"0ω - 0 "." "05""cos"2ω+0 "." "06""cos"4ω - 0 "." "11""cos"6ω+0 "." "32""cos"8ω} {} # matrix { {} # {} # {} } - 0 "." 5"cos"9ω+0 "." "32""cos""10"ω - 0 "." "11""cos""12"ω+0 "." "06""cos""14"ω {} # matrix { {} # {} # {} } - 0 "." "05""cos""16"ω+0 "." "04""cos""18"ω {} } } {}

H I ( ω ) = 0 . 04 sin + 0 . 05 sin 0 . 06 sin + 0 . 11 sin 0 . 32 sin + 0 . 5 sin 0 . 32 sin 10 ω + 0 . 11 sin 12 ω 0 . 06 sin 14 ω + 0 . 05 sin 16 ω 0 . 04 sin 18 ω H I ( ω ) = 0 . 04 sin + 0 . 05 sin 0 . 06 sin + 0 . 11 sin 0 . 32 sin + 0 . 5 sin 0 . 32 sin 10 ω + 0 . 11 sin 12 ω 0 . 06 sin 14 ω + 0 . 05 sin 16 ω 0 . 04 sin 18 ω alignl { stack { size 12{H rSub { size 8{I} } $$ω$$ = - 0 "." "04""sin"0ω+0 "." "05""sin"2ω - 0 "." "06""sin"4ω+0 "." "11""sin"6ω - 0 "." "32""sin"8ω} {} # matrix { {} # {} # {} } +0 "." 5"sin"9ω - 0 "." "32""sin""10"ω+0 "." "11""sin""12"ω - 0 "." "06""sin""14"ω {} # matrix { {} # {} # {} } +0 "." "05""sin""16"ω - 0 "." "04""sin""18"ω {} } } {}

These two components lead to the magnitude and phase spectra:

H ( ω ) = H R ( ω ) + H I ( ω ) H ( ω ) = H R ( ω ) + H I ( ω ) size 12{ lline H $$ω$$ rline = sqrt {H rSub { size 8{R} } $$ω$$ +H rSub { size 8{I} } $$ω$$ } } {}

Φ ( ω ) = tan 1 H I ( ω ) H R ( ω ) Φ ( ω ) = tan 1 H I ( ω ) H R ( ω ) size 12{Φ $$ω$$ ="tan" rSup { size 8{ - 1} } { {H rSub { size 8{I} } $$ω$$ } over {H rSub { size 8{R} } $$ω$$ } } } {}

The Matlab spectra are shown if Fig.3.32. We can see the highpass characterstic of the filter with transition frequency at π/2π/2 size 12{ {π} slash {2} } {}.

Example 3.7.5

Given the difference equation of a filter as

y ( n ) = 1 . 77 y ( n 1 ) 1 . 19 y ( n 2 ) + 0 . 28 y ( n 3 ) + 0 . 14 x ( n 2 ) y ( n ) = 1 . 77 y ( n 1 ) 1 . 19 y ( n 2 ) + 0 . 28 y ( n 3 ) + 0 . 14 x ( n 2 ) size 12{y $$n$$ =1 "." "77"y $$n - 1$$ - 1 "." "19"y $$n - 2$$ +0 "." "28"y $$n - 3$$ +0 "." "14"x $$n - 2$$ } {}

Find its magnitude and phase frequency spectrum.

Solution

Notice that the filter is IIR of third order. From the difference equation we can write directly the frequency response:

H ( ω ) = 0 . 14 e jω2 1 1 . 77 e + 1 . 19 e jω2 0 . 28 e jω3 H ( ω ) = 0 . 14 e jω2 1 1 . 77 e + 1 . 19 e jω2 0 . 28 e jω3 size 12{H $$ω$$ = { {0 "." "14"e rSup { size 8{ - jω2} } } over {1 - 1 "." "77"e rSup { size 8{ - jω} } +1 "." "19"e rSup { size 8{ - jω2} } - 0 "." "28"e rSup { size 8{ - jω3} } } } } {}

Matlab gives the magnitude and phase spectra depicted in Fig.3.33. Notice that with just a few terns the IIR gives a rather good smooth magnitude spectrum (refer to previous example of FIR filter).

Example 3.7.6

The following is the equation of a notch filter:

y ( n ) = 1 . 8523 y ( n 1 ) 0 . 94833 y ( n 2 ) x ( n ) 1 . 9021 x ( n 1 ) + x ( n 2 ) y ( n ) = 1 . 8523 y ( n 1 ) 0 . 94833 y ( n 2 ) x ( n ) 1 . 9021 x ( n 1 ) + x ( n 2 ) size 12{y $$n$$ =1 "." "8523"y $$n - 1$$ - 0 "." "94833"y $$n - 2$$ x $$n$$ - 1 "." "9021"x $$n - 1$$ +x $$n - 2$$ } {}

Plot its magnitude and phase spectra.

Solution

The frequency response is

H ( ω ) = 1 1 . 9021 e + e j2ω 1 1 . 8523 e + 0 . 94833 e j2ω H ( ω ) = 1 1 . 9021 e + e j2ω 1 1 . 8523 e + 0 . 94833 e j2ω size 12{H $$ω$$ = { {1 - 1 "." "9021"e rSup { size 8{ - jω} } +e rSup { size 8{ - j2ω} } } over {1 - 1 "." "8523"e rSup { size 8{ - jω} } +0 "." "94833"e rSup { size 8{ - j2ω} } } } } {}

Fig.3.34 shows the required spectra . The magnitude response gives a clear notes at frequency ω=0.ω=0. size 12{ω=0 "." 1π} {} rad/sample

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