Frequency-Domain C/S of LTI Systems

I LTI: Linear Time-Invariant system

I h(n), the of an LTI systems describes the time domain c/s.

I H(ω), the response describes the frequency-domain c/s. F I h(n) ←−−→ H(ω)

I study: system response to excitation signals that are a weighted linear combination of sinusoids or complex exponentials.

1/ 25 Frequency-Domain C/S of LTI Systems

I Recall the response of an LTI system to input signal x(n)

∞ X y(n) = h(n)x(n − k) (1) k=−∞

2/ 25 Frequency-Domain C/S of LTI Systems

I Excite the system with a complex exponential, i.e. let x(n) = Aejωn, −∞ < n < ∞, −π < ω < π

∞ X h i y(n) = h(k) Aejω(n−k) k=−∞ " ∞ # X (2) = A h(k)e−jωk ejωn k=−∞ = AejωnH(ω)

where, ∞ X H(ω) = h(k)e−jωk (3) k=−∞

3/ 25 Frequency-Domain C/S of LTI Systems

Observations

I y(n) is in the form of a complex exponential with same frequency as input, multiplied by a factor.

I The complex exponential signal x(n) is called an eigenfunction of the system.

I H(ω) evaluated at the frequency of the input is the corresponding eigenvalue of the system.

4/ 25 Frequency-Domain C/S of LTI Systems

Observations, cont.

I As H(ω) is a , it is periodic with period 2π

I H(ω) is a complex-valued function, can be expressed in polar form H(ω) = |H(ω)|eφω

I h(k) is related to H(ω) through

π 1 Z h(k) = H(ω)ejωk dω 2π −π

5/ 25 Frequency-Domain C/S of LTI Systems

Real-valued impulse response An LTI system with a real-valued impulse response, exhibits symmetry properties as derived in section 4.4.1

I |H(ω)| ⇒ even function of ω

I φ(ω) ⇒ odd function of ω.

I Consequently, it is enough to know |H(ω)| and φ(ω) for −π ≤ ω ≤ π,

6/ 25 Example

I Determine the output sequence of the system with impulse response 1 h(n) = ( )nu(n) 2 when the input signal is

x(n) = Aejπn/2, −∞ < n < ∞

7/ 25 Example, cont.

I Solution: evaluate H(ω)

∞ X 1 H(ω) = h(n)e−jωn = 1 −jω n=−∞ 1 − 2 e π 1 2 o H(ω = ) = = √ e−j26.6 2 1 1 + j 2 5

I Thus, output is   2 o y(n) = A √ e−j26.6 ejπn/2 5 2 o = √ Aej(πn/2−26.6 ), −∞ < n < ∞ 5

I system effect on the input: scale and shift

I change frequency, we change amount of scale change and phase shift. 8/ 25 Example: Moving Average Filter

I Determine the magnitude and phase of H(ω) for the three-point moving average (MA) system.

1 y(n) = [x(n + 1) + x(n) + x(n − 1)] 3 and plot these functions for 0 ≤ ω ≤ π

9/ 25 Example: Moving Average Filter, cont.

I Solution: 1 1 1 h(n) = { , , } 3 3 3 consequently,

X 1 1 H(ω) = h(k)e−jωk = (ejω + 1 + e−jω) = (1 + 2 cos ω) 3 3 n Hence 1 |H(ω)| = |1 + cos ω| 3  2π  0, 0 ≤ ω ≤ φ(ω) = 3 2π  π, ≤ ω < π 3

10/ 25 11/ 25 Example: system response to sinusoids

I Determine the response of the system in Example 5.1.1, for the input signal π x(n) = 10 − 5 sin n + 20 cos πn, −∞ < n < ∞ 2

I Recall, the frequency response of the system is 1 H(ω) = 1 −jω 1 − 2 e

12/ 25 Example, cont.

I Solution

I Idea Recognise the frequency of each part of the input signal, and find corresponding system response.

I First term 10, fixed signal ⇒ ω = 0 1 H(0) = 1 = 2 1 − 2 π I 5 sin 2 n has a frequency ω = π/2, thus

π 2 o H( ) = √ e−j26.6 2 5

I 20 cos πn has a frequency ω = π 2 H(π) = 3

13/ 25 c/s of LTI systems The General Case

I Most general case: input to the system is an arbitrary linear combination of sinusoids of the form

L X x(n) = Ai cos(ωi n + φi ), −∞ < n < ∞ i=1

Where {Ai } and {φi } amplitude and phase of corresponding sinusoidal component i .

I System response will be of the form:

L X y(n) = Ai |H(ωi )| cos [ωi n + φi + Θ(ωi )] i=1

Where |H(ωi )| and Θ(ωi ) are the magnitude and phase imparted by the system to the individual frequency

components of the input signal. 14/ 25 Steady-State and Transient Response to Sinusoidal Input Signals

I If excitation signal (exponential or sinusoidal) applied at some finite time instant,e.g. n = 0

response = steady-state + transient

I Example: let y(n) = ay(n − 1) + x(n) system response to any x(n) applied at n = 0

n X y(n) = an+1y(−1)+ ak x(n−k), n ≥ 0, y(−1) initial condition k=0

I Let x(n) be a complex exponential

x(n) = Aejωn, n ≥ 0

15/ 25 Steady-State and Transient Response to Sinusoidal Input Signals, cont.

I We get,

n X y(n) = an+1y(−1) + A ak ejω(n−k) k=0 " n # X = an+1y(−1) + A (ae−jω)k ejωn k=0 1 − an+1e−jω(n+1) = an+1y(−1) + A ejωn, n ≥ 0 1 − ae−jω Aan+1e−jω(n+1) A = an+1y(−1) − ejωn + ejωn 1 − ae−jω 1 − ae−jω

16/ 25 Steady-State and Transient Response to Sinusoidal Input Signals, cont.

I BIBO stable, if |a| < 1 n+1 I Since |a| < 1, the terms containing a → 0, as n → ∞.

I Steady-state response:

A jωn yss (n) = lim y(n) = e n→∞ 1 − ae−jω = AH(ω)ejωn

I Transient response of the system:

Aan+1e−jω(n+1) y (n) = an+1y(−1) − ejωn, n ≥ 0 tr 1 − ae−jω

17/ 25 Steady-State Response to Periodic Input Signals

I Let the input signal x(n) to a stable LTI be periodic with fundamental period N.

I Periodic → −∞ < n < ∞ → total response of the system is the steady-state response.

I Using the Fourier series representation of a periodic signal

N−1 X j2πkn/N x(n) = ck e , k = 0, 0, ..., N − 1 k=0

18/ 25 Steady-State Response to Periodic Input Signals, cont.

I Evaluating the system response for each complex exponential

j2πkn/N xk (n) = ck e , k = 0, 1, ..., N − 1 2πk  y (n) = c H ej2πkn/N , k = 0, 1, ..., N − 1 k k N where 2πk H( ) = H(ω)| , k = 0, 1, ..., N − 1 N ω=2πkn/N

19/ 25 Steady-State Response to Periodic Input Signals, cont.

I Superposition principle for linear systems:

N−1 X 2πk  y(n) = c H ej2πkn/N , −∞ < n < ∞ k N k=0

I LTI system response to a periodic input signal is also periodic with the same period N, with coefficients related by

2πk  d = c H , k = 0, 1, ..., N − 1 k k N

20/ 25 Response to Aperiodic Input Signals

I Let {x(n)} ne the aperiodic input sequence, {y(n)} output sequence, and {h(n)} unit sample response.

I By Convolution theorem

Y (ω) = H(ω)X (ω) In polar form, magnitude and phase of the output signal: |Y (ω)| = |H(ω)||X (ω)| ∠Y (ω) = ∠H(ω) + ∠X (ω)

I Output of an LTI system can NOT contain frequency components that are not contained in the input signal.

21/ 25 Response to Aperiodic Input Signals, cont.

I Energy density spectra of input and output

|Y (ω)|2 = |H(ω)|2|X (ω)|2 2 Syy (ω) = |H(ω)| Sxx (ω)

I Energy of the output signal

π 1 Z E = S (ω)dω y 2π yy −π π 1 Z = |H(ω)|2S (ω)dω 2π xx −π

22/ 25 Example

I A linear time-invariant system is characterized by its impulse response 12 h(n) = u(n) 2

I Determine the spectrum and energy density spectrum of the output signal when the system is excited by the signal

12 x(n) = u(n) 4

23/ 25 Example: Solution

I Frequency response of the system

∞ 2 X 1 H(ω) = e−jωn 2 n=0 1 = 1 −jω 1 − 2 e

I Similiarly, Fourier transform of the input sequence 1 X (ω) = 1 −jω 1 − 4 e

I Spectrum of the output signal

Y (ω) = H(ω)X (ω) 1 = 1 −jω 1 −jω (1 − 2 e )(1 − 4 e )

24/ 25 Example, cont.

I energy density spectrum

2 2 2 Sxx (ω) = |Y (ω)| = |H(ω)| |X (ω)| 1 = 5 17 1 ( 4 − cos ω)( 16 − 2 cos ω)

25/ 25