Frequency-Domain C/S of LTI Systems x(n) LTI y(n) LTI: Linear Time-Invariant system h(n), the impulse response of an LTI systems describes the time domain c/s. H(ω), the frequency response describes the frequency-domain c/s. h(n) F H(ω) 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 Recall the response of an LTI system to input signal x(n) y(n) = h(n)x(n k) (1) k= 2/ 25
Frequency-Domain C/S of LTI Systems Excite the system with a complex exponential, i.e. let x(n) = Ae jωn, < n <, π < ω < π y(n) = = A k= [ h(k) [Ae jω(n k)] k= = Ae jωn H(ω) h(k)e jωk ] e jωn (2) where, H(ω) = h(k)e jωk (3) k= 3/ 25
Frequency-Domain C/S of LTI Systems Observations y(n) is in the form of a complex exponential with same frequency as input, multiplied by a factor. The complex exponential signal x(n) is called an eigenfunction of the system. 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. As H(ω) is a Fourier transform, it is periodic with period 2π H(ω) is a complex-valued function, can be expressed in polar form H(ω) = H(ω) e φω h(k) is related to H(ω) through h(k) = 1 2π π π H(ω)e jωk dω 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 H(ω) even function of ω φ(ω) odd function of ω. Consequently, it is enough to know H(ω) and φ(ω) for π ω π, 6/ 25
Example Determine the output sequence of the system with impulse response when the input signal is h(n) = ( 1 2 )n u(n) x(n) = Ae jπn/2, < n < 7/ 25
Example, cont. Solution: evaluate H(ω) H(ω) = n= H(ω = π 2 ) = 1 1 + j 1 2 h(n)e jωn 1 = 1 1 2 e jω = 2 5 e j26.6o Thus, output is ( ) 2 y(n) = A 5 e j26.6o e jπn/2 = 2 5 Ae j(πn/2 26.6o), < n < system effect on the input: amplitude scale and phase shift change frequency, we change amount of scale change and phase shift. 8/ 25
Example: Moving Average Filter Determine the magnitude and phase of H(ω) for the three-point moving average (MA) system. y(n) = 1 [x(n + 1) + x(n) + x(n 1)] 3 and plot these functions for 0 ω π 9/ 25
Example: Moving Average Filter, cont. Solution: h(n) = { 1 3, 1 3, 1 3 } consequently, H(ω) = n h(k)e jωk = 1 3 (ejω + 1 + e jω ) = 1 (1 + 2 cos ω) 3 Hence H(ω) = 1 1 + cos ω 3 0, 0 ω 2π φ(ω) = 3 π, 2π 3 ω < π 10/ 25
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Example: system response to sinusoids Determine the response of the system in Example 5.1.1, for the input signal x(n) = 10 5 sin π n + 20 cos πn, < n < 2 Recall, the frequency response of the system is 1 H(ω) = 1 1 2 e jω 12/ 25
Example, cont. Solution Idea Recognise the frequency of each part of the input signal, and find corresponding system response. First term 10, fixed signal ω = 0 H(0) = 1 1 1 2 = 2 5 sin π 2 n has a frequency ω = π/2, thus H( π 2 ) = 2 5 e j26.6o 20 cos πn has a frequency ω = π H(π) = 2 3 13/ 25
Frequency Domain c/s of LTI systems The General Case Most general case: input to the system is an arbitrary linear combination of sinusoids of the form x(n) = L A i cos(ω i n + φ i ), < n < i=1 Where {A i } and {φ i } amplitude and phase of corresponding sinusoidal component i. System response will be of the form: y(n) = L A i H(ω i ) cos [ω i n + φ i + Θ(ω i )] i=1 Where H(ω i ) and Θ(ω i ) 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 If excitation signal (exponential or sinusoidal) applied at some finite time instant,e.g. n = 0 Example: let response = steady-state + transient y(n) = ay(n 1) + x(n) system response to any x(n) applied at n = 0 y(n) = a n+1 y( 1)+ n a k x(n k), n 0, y( 1) initial condition k=0 Let x(n) be a complex exponential x(n) = Ae jωn, n 0 15/ 25
Steady-State and Transient Response to Sinusoidal Input Signals, cont. We get, y(n) = a n+1 y( 1) + A n a k e jω(n k) k=0 [ n ] = a n+1 y( 1) + A (ae jω ) k k=0 e jωn = a n+1 y( 1) + A 1 an+1 e jω(n+1) 1 ae jω e jωn, n 0 = a n+1 y( 1) Aan+1 e jω(n+1) 1 ae jω e jωn A + ejωn 1 ae jω 16/ 25
Steady-State and Transient Response to Sinusoidal Input Signals, cont. BIBO stable, if a < 1 Since a < 1, the terms containing a n+1 0, as n. Steady-state response: y ss (n) = lim n y(n) = = AH(ω)e jωn Transient response of the system: A ejωn 1 ae jω y tr (n) = a n+1 y( 1) Aan+1 e jω(n+1) 1 ae jω e jωn, n 0 17/ 25
Steady-State Response to Periodic Input Signals Let the input signal x(n) to a stable LTI be periodic with fundamental period N. Periodic < n < total response of the system is the steady-state response. Using the Fourier series representation of a periodic signal x(n) = N 1 k=0 c k e j2πkn/n, k = 0, 0,..., N 1 18/ 25
Steady-State Response to Periodic Input Signals, cont. Evaluating the system response for each complex exponential x k (n) = c k e j2πkn/n, k = 0, 1,..., N 1 ( ) 2πk y k (n) = c k H e j2πkn/n, k = 0, 1,..., N 1 N where H( 2πk N ) = H(ω) ω=2πkn/n, k = 0, 1,..., N 1 19/ 25
Steady-State Response to Periodic Input Signals, cont. Superposition principle for linear systems: y(n) = N 1 k=0 c k H ( ) 2πk e j2πkn/n, N < n < LTI system response to a periodic input signal is also periodic with the same period N, with coefficients related by ( ) 2πk d k = c k H, k = 0, 1,..., N 1 N 20/ 25
Response to Aperiodic Input Signals Let {x(n)} ne the aperiodic input sequence, {y(n)} output sequence, and {h(n)} unit sample response. 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) 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. Energy density spectra of input and output Y (ω) 2 = H(ω) 2 X (ω) 2 S yy (ω) = H(ω) 2 S xx (ω) Energy of the output signal E y = 1 2π = 1 2π π π π π S yy (ω)dω H(ω) 2 S xx (ω)dω 22/ 25
Example A linear time-invariant system is characterized by its impulse response ( ) 1 2 h(n) = u(n) 2 Determine the spectrum and energy density spectrum of the output signal when the system is excited by the signal x(n) = ( ) 1 2 u(n) 4 23/ 25
Example: Solution Frequency response of the system H(ω) = n=0 ( ) 1 2 e jωn 2 1 = 1 1 2 e jω Similiarly, Fourier transform of the input sequence X (ω) = 1 1 1 4 e jω Spectrum of the output signal Y (ω) = H(ω)X (ω) 1 = (1 1 2 e jω )(1 1 4 e jω ) 24/ 25
Example, cont. energy density spectrum S xx (ω) = Y (ω) 2 = H(ω) 2 X (ω) 2 = ( 5 4 1 17 cos ω)( 16 1 2 cos ω) 25/ 25