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FREQUENCY RESPONSE OF LTI (LSI) SYSTEMS

Module by: Nguyen Huu Phuong

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
H(ω)= ∑ n=−∞ ∞ h(n) e −jωn H(ω)= ∑ n=−∞ ∞ h(n) e −jωn MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIeacaGGOaGaeqyYdCNaaiykaiabg2da9maaqahabaGaamiAaiaacIcacaWGUbGaaiykaiaadwgadaahaaWcbeqaaiabgkHiTiaadQgacqaHjpWDcaWGUbaaaaqaaiaad6gacqGH9aqpcqGHsislcqGHEisPaeaacqGHEisPa0GaeyyeIuoaaaa@4BC9@ (1)
And the inverse DTFT is
h(n)= 1 2π ∫ −π π H(ω) e jωn dω h(n)= 1 2π ∫ −π π H(ω) e jωn dω MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIgacaGGOaGaamOBaiaacMcacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaGaeqiWdahaamaapedabaGaamisaiaacIcacqaHjpWDcaGGPaaaleaacqGHsislcqaHapaCaeaacqaHapaCa0Gaey4kIipakiaadwgadaahaaWcbeqaaiaadQgacqaHjpWDcaWGUbaaaOGaamizaiabeM8a3baa@4F7B@ (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) :
y(n)=x(n)*h(n)↔Y(ω)=X(ω)H(ω) y(n)=x(n)*h(n)↔Y(ω)=X(ω)H(ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMhacaGGOaGaamOBaiaacMcacqGH9aqpcaWG4bGaaiikaiaad6gacaGGPaGaaiOkaiaadIgacaGGOaGaamOBaiaacMcacqGHugYQcaWGzbGaaiikaiabeM8a3jaacMcacqGH9aqpcaWGybGaaiikaiabeM8a3jaacMcacaWGibGaaiikaiabeM8a3jaacMcaaaa@5051@ (3)
Or
H(ω)= Y(ω) X(ω) H(ω)= Y(ω) X(ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIeacaGGOaGaeqyYdCNaaiykaiabg2da9maalaaabaGaamywaiaacIcacqaHjpWDcaGGPaaabaGaamiwaiaacIcacqaHjpWDcaGGPaaaaaaa@42F5@ (4)
Figure 1: Maping time domain to frequency domain using the time convolution property
The frequency response H(ω) H(ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaaiaacaqaaeaadaqaaqaaaOqaaiaadIeacaGGOaGaeqyYdCNaaiykaaaa@39D5@ is usually a complex quantily, so we write
H(ω)= H R (ω)+j H 1 (ω)=| H(ω) | e jΦ(ω) H(ω)= H R (ω)+j H 1 (ω)=| H(ω) | e jΦ(ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIeacaGGOaGaeqyYdCNaaiykaiabg2da9iaadIeadaWgaaWcbaGaamOuaaqabaGccaGGOaGaeqyYdCNaaiykaiabgUcaRiaadQgacaWGibWaaSbaaSqaaiaaigdaaeqaaOGaaiikaiabeM8a3jaacMcacqGH9aqpdaabdaqaaiaadIeacaGGOaGaeqyYdCNaaiykaaGaay5bSlaawIa7aiaadwgadaahaaWcbeqaaiaadQgacqqHMoGrcaGGOaGaeqyYdCNaaiykaaaaaaa@5554@ (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
Φ(ω)= tan⁡ −1 H I (ω) H R (ω) Φ(ω)= tan⁡ −1 H I (ω) H R (ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfA6agjaacIcacqaHjpWDcaGGPaGaeyypa0JaciiDaiaacggacaGGUbWaaWbaaSqabeaacqGHsislcaaIXaaaaOWaaSaaaeaacaWGibWaaSbaaSqaaiaadMeaaeqaaOGaaiikaiabeM8a3jaacMcaaeaacaWGibWaaSbaaSqaaiaadkfaaeqaaOGaaiikaiabeM8a3jaacMcaaaaaaa@4A42@ (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),
| H(−ω) |=| H(ω) | and Φ(−ω)=−Φ(ω) | H(−ω) |=| H(ω) | and Φ(−ω)=−Φ(ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaamisaiaacIcacqGHsislcqaHjpWDcaGGPaaacaGLhWUaayjcSdGaeyypa0ZaaqWaaeaacaWGibGaaiikaiabeM8a3jaacMcaaiaawEa7caGLiWoacaaMf8Uaamyyaiaad6gacaWGKbGaaGzbVlabfA6agjaacIcacqGHsislcqaHjpWDcaGGPaGaeyypa0JaeyOeI0IaeuOPdyKaaiikaiabeM8a3jaacMcaaaa@5800@ (8)
The frequency response of a system exists if the system is BIBO stable, i.e. (Equation)
∑ n=−∞ ∞ | h(n) | <∞ ∑ n=−∞ ∞ | h(n) | <∞ MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaamaaqahabaWaaqWaaeaacaWGObGaaiikaiaad6gacaGGPaaacaGLhWUaayjcSdaaleaacaWGUbGaeyypa0JaeyOeI0IaeyOhIukabaGaeyOhIukaniabggHiLdGccqGH8aapcqGHEisPaaa@46C9@ (9)
Example 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) ] MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadMhacaGGOaGaamOBaiaacMcacqGH9aqpcaWGbbWaamWaaeaacaWG4bGaaiikaiaad6gacqGHsislcaaIYaGaaiykaiabgUcaRiaadIhacaGGOaGaamOBaiabgkHiTiaaigdacaGGPaGaey4kaSIaamiEaiaacIcacaWGUbGaaiykaiabgUcaRiaadIhacaGGOaGaamOBaiabgUcaRiaaigdacaGGPaGaey4kaSIaamiEaiaacIcacaWGUbGaey4kaSIaaGOmaiaacMcaaiaawUfacaGLDbaaaaa@576E@
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
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) ] MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIgacaGGOaGaamOBaiaacMcacqGH9aqpcaWGbbWaamWaaeaacqaH0oazcaGGOaGaamOBaiabgkHiTiaaikdacaGGPaGaey4kaSIaeqiTdqMaaiikaiaad6gacqGHsislcaaIXaGaaiykaiabgUcaRiabes7aKjaacIcacaWGUbGaaiykaiabgUcaRiabes7aKjaacIcacaWGUbGaey4kaSIaaGymaiaacMcacqGHRaWkcqaH0oazcaGGOaGaamOBaiabgUcaRiaaikdacaGGPaaacaGLBbGaayzxaaaaaa@5AA5@
It turns out that this impulse response is the same as the signal in Example , hence the frequency response of the system is
H(ω)= ∑ n=−2 2 h(n) e −jωn =A(1+2cos⁡ω+2cos⁡2ω) H(ω)= ∑ n=−2 2 h(n) e −jωn =A(1+2cos⁡ω+2cos⁡2ω) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIeacaGGOaGaeqyYdCNaaiykaiabg2da9maaqahabaGaamiAaiaacIcacaWGUbGaaiykaiaadwgadaahaaWcbeqaaiabgkHiTiaadQgacqaHjpWDcaWGUbaaaaqaaiaad6gacqGH9aqpcqGHsislcaaIYaaabaGaaGOmaaqdcqGHris5aOGaeyypa0JaamyqaiaacIcacaaIXaGaey4kaSIaaGOmaiGacogacaGGVbGaai4CaiabeM8a3jabgUcaRiaaikdaciGGJbGaai4BaiaacohacaaIYaGaeqyYdCNaaiykaaaa@5B81@
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
H(ω)= ∑ n=−∞ ∞ h(n) e −jωn = ∑ n=0 ∞ (0.8 e −jω ) n = 1 1−0.8 e −jω H(ω)= ∑ n=−∞ ∞ h(n) e −jωn = ∑ n=0 ∞ (0.8 e −jω ) n = 1 1−0.8 e −jω MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@6716@
In order to compute the real and imaginary frequency responses we write
H(ω)= 1−0.8 e jω (1−0.8 e −jω )(1−0.8 e jω ) = 1−0.8cos⁡ω−j0.8sin⁡ω 1.64−1.6cos⁡ω H(ω)= 1−0.8 e jω (1−0.8 e −jω )(1−0.8 e jω ) = 1−0.8cos⁡ω−j0.8sin⁡ω 1.64−1.6cos⁡ω MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@7286@
From this,
H R (ω)= 1−0.8cos⁡ω 1.64−1.6cos⁡ω H R (ω)= 1−0.8cos⁡ω 1.64−1.6cos⁡ω MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIeadaWgaaWcbaGaamOuaaqabaGccaGGOaGaeqyYdCNaaiykaiabg2da9maalaaabaGaaGymaiabgkHiTiaaicdacaGGUaGaaGioaiGacogacaGGVbGaai4CaiabeM8a3bqaaiaaigdacaGGUaGaaGOnaiaaisdacqGHsislcaaIXaGaaiOlaiaaiAdaciGGJbGaai4BaiaacohacqaHjpWDaaaaaa@4F16@
H I (ω)= 0.8sin⁡ω 1.64−1.6cos⁡ω H I (ω)= 0.8sin⁡ω 1.64−1.6cos⁡ω MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadIeadaWgaaWcbaGaamysaaqabaGccaGGOaGaeqyYdCNaaiykaiabg2da9maalaaabaGaaGimaiaac6cacaaI4aGaci4CaiaacMgacaGGUbGaeqyYdChabaGaaGymaiaac6cacaaI2aGaaGinaiabgkHiTiaaigdacaGGUaGaaGOnaiGacogacaGGVbGaai4CaiabeM8a3baaaaa@4D6A@
Figure 2: Example 2
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).
Figure 3: Example 3(frequency response of an ideal lowpass filter)
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:
h(n)= 1 n ∫ −∞ ∞ e jωn dω= 1 2π ∫ − ω c ω c e jωn dω  = 1 2π e jωn jn | − ω c ω c                    = sin⁡ ω c n πn                   = ω c π sin⁡ ω c n ω c n h(n)= 1 n ∫ −∞ ∞ e jωn dω= 1 2π ∫ − ω c ω c e jωn dω  = 1 2π e jωn jn | − ω c ω c                    = sin⁡ ω c n πn                   = ω c π sin⁡ ω c n ω c n MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@C870@
The result can be left in either of the two forms above. In the latter form the result contains the sin⁡x /x sin⁡x /x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaalyaabaGaci4CaiaacMgacaGGUbGaamiEaaqaaiaadIhaaaaaaa@3AC6@ function (section) whose limit as x→0 x→0 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 sin⁡x /x sin⁡x /x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaalyaabaGaci4CaiaacMgacaGGUbGaamiEaaqaaiaadIhaaaaaaa@3AC6@ function and taking the limit as x→0 x→0 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 (sin⁡n ω c ) d dn (nπ) | n=0 =  ω c cos⁡n ω c π | n=0 =  ω c π h(n)= d dn (sin⁡n ω c ) d dn (nπ) | n=0 =  ω c cos⁡n 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
h(n)= sin⁡ ω c n πn  ,  n≠0     ω c π  ,     n=0 h(n)= sin⁡ ω c n πn  ,  n≠0     ω c π  ,     n=0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOabaeqabaGaamiAaiaacIcacaWGUbGaaiykaiabg2da9maalaaabaGaci4CaiaacMgacaGGUbGaeqyYdC3aaSbaaSqaaiaadogaaeqaaOGaamOBaaqaaiabec8aWjaad6gaaaGaaGjbVlaacYcacaaMf8UaaGzbVlaad6gacqGHGjsUcaaIWaaabaGaaGzbVlaaywW7caaMf8+aaSaaaeaacqaHjpWDdaWgaaWcbaGaam4yaaqabaaakeaacqaHapaCaaGaaGjbVlaacYcacaaMf8UaaGzbVlaaywW7caaMe8UaaGjbVlaad6gacqGH9aqpcaaIWaaaaaa@627D@
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@ .
Figure 4: Example 3 (impulse response of ideal lowpass filter for various 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=0∣H(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.
Figure 5:
Figure 6: | H(ω) | dB | H(ω) | dB MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaamisaiaacIcacqaHjpWDcaGGPaaacaGLhWUaayjcSdWaaSbaaSqaaiaadsgacaWGcbaabeaaaaa@3ED6@ verus | H(ω) | | H(ω) | MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=vqpWqadeaabiGaciaacaqabeaadaqaaqaaaOqaamaaemaabaGaamisaiaacIcacqaHjpWDcaGGPaaacaGLhWUaayjcSdaaaa@3CFA@
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)∣dB∣H(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)∣dB∣H(ω)∣dB size 12{ lline H \( ω \) rline rSub { size 8{ ital "dB"} } } {}.
Figure 7: Example of magnitude response in linear and 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)=