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# FREQUENCY RESPONSE OF LTI (LSI) SYSTEMS

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

## FREQUENCY RESPONSE OF LTI (LSI) SYSTEMS

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

(1)

And the inverse DTFT is

(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) in time domain to its transform Y(ω)Y(ω) size 12{Y $$ω$$ } {} in the frequency domain (Figure ) :

(3)

Or

(4)

(5)

where

| H(ω) |= H R 2 (ω)+ H I 2 (ω) | H(ω) |= H R 2 (ω)+ H I 2 (ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaamisaiaacIcacqaHjpWDcaGGPaaacaGLhWUaayjcSdGaeyypa0ZaaOaaaeaacaWGibWaa0baaSqaaiaadkfaaeaacaaIYaaaaOGaaiikaiabeM8a3jaacMcacqGHRaWkcaWGibWaa0baaSqaaiaadMeaaeaacaaIYaaaaOGaaiikaiabeM8a3jaacMcaaSqabaaaaa@4A6E@
(6)

and

(7)

are, respectively, the magnitude response and the phase response. If the impulse response h(n) is real-valued then, as for DTFT of signal ( Equation ),

(8)

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

(9)

### Example 1

where A is a constant.

Solution

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

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

The system is a low-pass filter.

### Example 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 . The frequency response is

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

From this,

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} } } {}

Figure 2 presents all the required spectra.

### Example 3

The frequence response of an ideal lowpass filter having cutoff frequence (Figure 3) is H(ω)=1, ω c ω ω c =0,otherwise H(ω)=1, ω c ω ω c =0,otherwise MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOabaeqabaGaamisaiaacIcacqaHjpWDcaGGPaGaeyypa0JaaGymaiaaywW7caGGSaGaaGzbVlaaywW7cqGHsislcqaHjpWDdaWgaaWcbaGaam4yaaqabaGccqGHKjYOcqaHjpWDcqGHKjYOcqaHjpWDdaWgaaWcbaGaam4yaaqabaaakeaacaaMf8UaaGzbVlaaysW7cqGH9aqpcaaIWaGaaGzbVlaacYcacaaMf8UaaGzbVlaad+gacaWG0bGaamiAaiaadwgacaWGYbGaam4DaiaadMgacaWGZbGaamyzaiaaywW7aaaa@62C5@ Find its impluse response h(n).

Solution

Recall that the frequency response of a digital system is periodic with a period of , with the central period taken as [ 0,2π ] [ 0,2π ] MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaadmaabaGaaGimaiaacYcacaaIYaGaeqiWdahacaGLBbGaayzxaaaaaa@3BB3@ or , more usually , [ π,π ] [ π,π ] MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaadmaabaGaeyOeI0IaeqiWdaNaaiilaiabec8aWbGaay5waiaaw2faaaaa@3CE7@ . The impulse response is the inverse DTFT of the frequency response:

The result can be left in either of the two forms above. In the latter form the result contains the sinx /x sinx /x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaalyaabaGaci4CaiaacMgacaGGUbGaamiEaaqaaiaadIhaaaaaaa@3AC6@ function ( section ) whose limit as x0 x0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIhacqGHsgIRcaaIWaaaaa@3982@ is 1.

We should treat the case n = 0 separately in one of the three ways : (1) replacing n = 0 in the initial integral and taking the integration , (2) put the result in terms of sinx /x sinx /x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaalyaabaGaci4CaiaacMgacaGGUbGaamiEaaqaaiaadIhaaaaaaa@3AC6@ function and taking the limit as x0 x0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaaiaadIhacqGHsgIRcaaIWaaaaa@3982@ , and (3) using L’Hospital’s rule which is

h(n)= d dn (sinn ω c ) d dn (nπ) | n=0 = ω c cosn ω c π | n=0 = ω c π h(n)= d dn (sinn ω c ) d dn (nπ) | n=0 = ω c cosn ω c π | n=0 = ω c π MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@6907@

Thus the impulse response is

Figure 4 shows the result for 4 different values of cutoff frequency ω c ω c MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3naaBaaaleaacaWGJbaabeaaaaa@38C4@ .

## 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 =20 log 10 | H(ω) | | H(ω) | dB =20 log 10 | H(ω) | MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaamisaiaacIcacqaHjpWDcaGGPaaacaGLhWUaayjcSdWaaSbaaSqaaiaadsgacaWGcbaabeaakiabg2da9iaaikdacaaIWaGaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaigdacaaIWaaabeaakmaaemaabaGaamisaiaacIcacqaHjpWDcaGGPaaacaGLhWUaayjcSdaaaa@4CEC@
(10)

See Figure 7. Remember when H(ω)H(ω) 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)H(ω) size 12{ lline H $$ω$$ rline } {}> 1 the dBs are positive, when H(ω)H(ω) 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. Figure 7 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

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

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

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 ω MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3baa@37B2@ ), so

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

### Example 4

Find the output for the input

Solution

First we find the filter frequency response

Notice the both signals in (a) and (b) have the same angular frequency of ω=π/2 ω=π/2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3jabg2da9maalyaabaGaeqiWdahabaGaaGOmaaaaaaa@3B47@ . 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

(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

## Frequency response of systems in cascade or in parallel

In various situations filters are connected in cascade or in parallel. Section has presented this matter with respect to system impulse responses. Now we treat the problem with respect to frequency responses. By using the associativity and the distributivity of impulse responses, and the convolution theorem of DTFT we can obtain (Figure 11):

(13)

Systems in parallel

(14)

## Frequency response in terms of filter coefficients

From the difference equation of general linear filter ( Equation )

(15)

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

We replace this into the difference equation:

Thus

(16)

Notice that if the signal difference equation is written differently (as some authors do) the above expression for H(ω) H(ω) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqipDI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaaiaacaqaaeaadaqaaqaaaOqaaKqzGfGaamisaiaacIcacqaHjpWDcaGGPaaaaa@3AC5@ does not apply.

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. Also notice that for nonrecursive filter, the denominator is just 1.

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

(17)

Where Φ N (ω) Φ N (ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfA6agnaaBaaaleaacaWGobaabeaakiaacIcacqaHjpWDcaGGPaaaaa@3B8E@ or N(ω) N(ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgcIiqlaaysW7caWGobGaaiikaiabeM8a3jaacMcaaaa@3D09@ is the phase of N(ω)N(ω) size 12{N $$ω$$ } {}, and Φ D (N)orD(ω) Φ D (N)orD(ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfA6agnaaBaaaleaacaWGebaabeaakiaacIcacaWGobGaaiykaiaaywW7caWGVbGaamOCaiaaywW7cqGHGic0caaMe8UaamiraiaacIcacqaHjpWDcaGGPaaaaa@46AB@ is the phase of D(ω)D(ω) size 12{D $$ω$$ } {}. The magnitude and phase responses are then given respectively by

(18)

(19)

### Example 5

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 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@9353@ Write the difference equation of the filter and find the frequency response.

Solution

The difference equation the FIR filter is

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

These two components lead to the magnitude and phase spectra:

The Matlab spectra are shown if Figure 9. We can see the highpass characterstic of the filter with transition frequency at π π MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabec8aWbaa@37A2@ /2.

### Example 6

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:

Matlab gives the magnitude and phase spectra depicted in Figure 10. Notice that with just a few terns the IIR gives a rather good smooth magnitude spectrum (refer to previous example of FIR filter).

### Example 7

Plot its magnitude and phase spectra.

Solution

The frequency response is

Figure 11shows 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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