Discrete-time Signals and Systems in

Discrete-time Signals and Systems in the Frequency Domain Chapter 3, Sections 3.1-39 3.9 Chapter 4, Sections 4.8-4.9 Dr. Iyad Jafar

Outline Introduction The Continuous-Time FourierTransform (CTFT) The Discrete-Time Fourier Transform (DTFT) DTFT Properties DTFT of Common Sequences DTFTTh Theorems Relation Between CTFT and DTFT Frequency Analysis of Discrete-Time Systems 2 The Concept of Filtering i

The Continuous-Time Fourier Transform (CTFT) For a continuous-time time aperiodic signal, x a (t), the CTFT (Fourier Spectrum or simply Spectrum) is defined as where Ω is the continuous angular frequency in radian/sec The continuous-time signal can be obtained from X a (j Ω)using A CTFT pair is denoted by 3

The Continuous-Time Fourier Transform (CTFT) The quantity X a (j Ω) is a complex function of Ω in the range <Ω< where 4 The quantity Xa( jω) is called the magnitude spectrum and the quantity θ a (Ω) is called the phase spectrum with both spectrums are real lfunctions of fωω In general, X a ( jω) exists if x a (t) has finite it discontinuities iti and finite it maxima and minima i in any finite interval x a a( (t) is absolutely integrable, i.e. x(t)dt< a

Finite-Dimensional LTI Discrete-Time Example 4.1: a) x(t) a (t) Systems j t X(j a ) (t) e dt j0 e = 1 By sifting property b) t x(t) a e u(t) 1 X(j a ) e e dt = e dt = e j t j t (j )t ( j ) 0 0 0 1 1 1 1 X(j a ) = e 1 j 2 2 jtan ( / ) 2 2 e jtan ( / ) 1 X(j a ) ( ) tan 1 ( ) 2 2 a 5

Finite-Dimensional LTI Discrete-Time Systems Example 4.1: Continued. b) 2 1.5 1.5 1 05 0.5 X a (j) 1 a (j) 0 0.5-0.5-1 6 0-5 0 5-1.5-5 0 5 1 X(j 1 a ) 2 2 a ( ) tan ( )

The Continuous-Time Fourier Transform (CTFT) For energy signals, which can be written as so, 7

The Continuous-Time Fourier Transform (CTFT) Hence, The above relation is more commonly known as the Parseval s relation for finite energy continuous-time signals.theterm X a ( jω) 2 is called the energy density spectrum S xx (Ω) 8

The Continuous-Time Fourier Transform (CTFT) Classification of signals based on the span of the spectrum Full-band d the spectrum is dfi definedd - < Ω < Band-limited the spectrum spans a finite range Ω 1 Ω Ω 2 Lowpass signals 0 Ω Ω p Highpass gp signals 0<Ω p Ω < Bandpass signals 0<Ω L Ω Ω H < 9

The Discrete-Time Fourier Transform (DTFT) The DTFT for an discrete-time signal x[n] that is obtained through sampling of an analog signal x a (t) with sampling period T is given by j j t X(e ) X a(j ) x a (t)e dt j X(e ) tnt x[n]e n j j nt X(e ) x a (nt)e dt jn where ω is the normalized angular frequency in radian/sample In general, X(e jω ) is a complex function of the real variable ω and can be written as 10

Example 4.2: The Discrete-Time Fourier Transform (DTFT) a) The DTFT of the unit sample sequence δ[n] is given by j j n j n X(e ) x[n]e [n]e 1 n n b) The DTFT of x[n] = {0, 1, 2,-1,0} 10} 4 j jn jn X(e ) x[n]e x[n]e n n0 X(e ) x[0]e x[1]e x[2]e x[3]e x[4]e j j0 j j2 j3 j4 X(e ) e 2e - e j j j2 j3 11 j X(e ) (cos + 2cos 2 - cos 3 ) j(sin + 2sin 2 - sin 3 )

The Discrete-Time Fourier Transform (DTFT) Example 4.2: Continued b) The DTFT of x[n] = {0, 1, 2,-1,0} X(e ) (cos + 2cos 2 - cos 3 ) + (sin + 2sin 2 - sin 3 ) j 2 2 3 2.8 2.6 X(e j ) 2.4 22 2.2 2 12 1.8-3 -2-1 0 1 2 3

The Discrete-Time Fourier Transform (DTFT) Example 4.2: Continued b) The DTFT of x[n] = {0, 1, 2,-1,0} -1 (sin +2sin 2 - sin 3 ) ( ) tan (cos + 2cos 2 - cos 3 ) 4 0-2 2-4 () 0 () -6-8 -2-10 -12 13-4 -4-2 0 2 4 Phase -14-4 -2 0 2 4 Unwrapped phase?!

The Discrete-Time Fourier Transform (DTFT) Example 4.3: Compute the DTFT for the sequence x[n] = α n u[n], α <1 14

The Discrete-Time Fourier Transform (DTFT) Example 4.4: 4: Compute the DTFT for the sequence x[n] A, 0 n L - 1 0, otherwise 15

The Discrete-Time Fourier Transform (DTFT) The inverse discrete-time time FT (IDTFT) is given by 1 j jn x[n] X(e )e d 2 DTFT j x[n] X(e ) Existence of DTFT The DTFT exist if the sequence x[n] is absolutely summable! 16 So, the DTFT Talways exists for finite-length sequences

DTFT Properties 1. The DTFT X(e jω ) is a periodic function with period of 2π. Both the magnitude and phase are periodic. j X(e ) jn j2k j( 2k)n x[n]e X(e ) x[n]e n n 17 j2k jn j2kn n X(e j2k jn j ) x[n]e e X(e ) x[n]e X(e ) n Remember also that the maximum rate of oscillation is π Full-band discrete-time signals - π < ω π Band-limited discrete-time signals Lowpass signals 0 ω ω p π Highpass signals 0<ω p ω π Bandpass signals 0<ω L ω ω H π

Example 4.5. x[n] = (0.5) n u[n] DTFT Properties 2 0.6 X(e j ) 1.8 1.6 1.4 1.2 1 0.8-9.42-6.28-3.14 0 3.14 6.28 9.42 () 0.4 0.2 0-0.2-0.4-0.6-0.8-9.42-6.28-3.14 0 3.14 6.28 9.42 18

DTFT Properties 2. For a complex-valued sequence x[n], the following symmetry properties hold 19

DTFT Properties 3. For a real-valued sequence x[n], the following symmetry properties hold 20

21 DTFT of Common Sequences

22 DTFT Theorems

DTFT Examples Example 4.6: Find the DTFT of the sequence n x[n], 0 n M-1, < 1 23

DTFT Examples Example 4.7: Find the DTFT Y(e jω ) if y[n] + 6y[n-1] [n] 3 [n-1] 24

DTFT Examples n n Example 4.8: Find the DTFT for x[n] = (-1) u[n], < 1 25

DTFT Examples Example 4.9: Find the DTFT for n y[n] = (n+1) u[n], < 1 26

Relation between CTFT and DTFT A discrete-time signal g[n] is obtained from an analog signal g a (t) throughsampling with sampling frequency Ω T =2πF T =2π/T g p g p g q y T T What is the relation between G(e jω ) and G a a(j Ω)? )??? The sampling process can be modeled as multiplying g a (t) by a periodic train of impulses p(t) with period T to obtain the sampled signal g p (t) 27

Relation between CTFT and DTFT The train of impulses p(t) is given by and its Fourier series expansion is and the CTFT of p(t) is 1 2k 1 P(j ) ( ) ( k T ) T T T k k According to the convolution theorem, the Fourier transform of g p (t) is basically the convolution between G a (j Ω) and P(j Ω) 28 p a G(j ) G(j ) P(j( )) d

Relation between CTFT and DTFT 1 G(j p ) G a(j ) ( k T ) d T k Interchanging the order of integration and summation 1 G(j p ) G(j a )( k T ) d T k by sifting property 1 G(j ) G (j( k )) p a T T k This implies that the sampled spectrum G p (jω) isaperiodic function formed from scaled replicas of G a (jω) placed at integer multiples of 29 the sampling frequency Ω T

Relation between CTFT and DTFT Also, note that So, the DTFT G(e jω ) is the same as G p (jω) with the frequency axis scaled using ω = ΩT. Now, if Ω = Ω T,thenω = Ω T T=2πT/T = 2π This implies that G(e jω ) is periodic with period of 2π. 30

Relation between CTFT and DTFT X = Scaling the frequency axis 31

Relation between CTFT and DTFT Aliasing and SamplingTheorem For a band-limited signal g a (t), i.e. G a (jω) is limited to Ω Ω M, that is sampled with a sampling rate Ω T that is sufficiently large, the spectrum will be The original signal g a (t) can be recovered from the sampled signal by g p (t) by multiplying G p (jω) withlowpassfilter. y g p y y g p 32 H LP (j ) T, - M 0, otherwise M

Relation between CTFT and DTFT Aliasing and SamplingTheorem However, if Ω M is too large or the sampling rate Ω T is too small, then the aliases or copies of G p (jω) will overlap. 33 This implies that the original signal g a (t) can not be recovered from the sampled signal by g p(t). To avoid this, Ω T - Ω M Ω M Ω T 2Ω M.This value 2Ω M for the sampling rate is called the Nyquist rate. SamplingTheorem Let g a (t) be a band-limited signal with G a (jω) =0, Ω Ω M, then g a (t) is uniquely determined d by its samples ga(nt), if Ω T 2Ω M.

Frequency Analysis of LTI Discrete Systems Most discrete-time time signals encountered in practice can be represented as a linear combination of a very large, maybe infinite, number of sinusoidal discrete-time time signals of different angular frequencies Thus, knowing the response of the LTI system to a single sinusoidal signal, we can determine its response to more complicated signals by making use of the superposition property. In the following, we investigate the behavior of LTI systems in response to input sinusoids 34

Frequency Analysis of LTI Discrete Systems Consider the LTI discrete-time time system with an impulse response h[n] shown below The output is defined as the convolution sum between x[n] and h[n] If we now assume that x[n] to be a complex sinusoid that was applied to the system at t = -, i.e. 35 j o n x[n] e, - < n <

Frequency Analysis of LTI Discrete Systems Accordingly, the output y[n] will be = 36 Hence, the output of the LTI system to a complex sinusoid with frequency ω o is a complex sinusoid scaled by H(e jωo ) and shifted by θ(ω o ) = arg{h(e jωo )}. The quantity H(e jωo )iscalledthefrequency response of the system and it is essentially the DTFT of the impulse response h[n] and it completely l describes the system in the frequency domain. H(e jωo ) is the magnitude response, and θ(ω o )isthe phase response

Frequency Analysis of LTI Discrete Systems Example 4.10: Let the impulse response of a LTI system be h[n] = {-3, 10, 3}, -1 n 1, what is y[n] if x[n] = 5 cos(0.2πn)? 37

Frequency Analysis of LTI Discrete Systems For an LTI system with impulse response h[n] y[n] x[nk]h[k] k Tki Taking the DTFT of fboth sides jn Y(e ) h[k] x[nk] e k n j n j n y[n] e x[n k]h[k] e n n k jn Interchanging the order of summation Y(e ) X(e ) h[k] e jn jn jk k jn jn jn Y(e ) X(e )H(e ) By time-shift property 38 j n j Y(e ) H(e ) X(e jn )

Frequency Analysis of LTI Discrete Systems For a FIR LTI system with impulse response h[n], N1 n N2 N2 jn jk H(e ) h[k] e nn 1 For an IIR LTI system described by a LCCDE N k k k0 k0 N M j jk j j k ay(e k ) e = b k X(e ) e k0 k0 M a y[n k] = b x[n - k] Polynomial in e jω 39 M j k b j k e Y(e ) j k 0 = H(e ) j N X(e ) j k a k e k 0 Rational Polynomial in e jω

Frequency Analysis of LTI Discrete Systems What if the input sinusoid is not finite in duration? Usually signals are applied at specific time to the system (Gating)? How is this related to the frequency response? Let x[n] = e jωon u[n], then the convolution with impulse response h[n] becomes which can be written o y[n] h[m] e h[m] e j (n-m) j (n-m) m mn1 o 40 y[n] H(e )e h[m] e e mn1 j j n -j m j n Steady State Response o o o o Transient Response

Frequency Analysis of LTI Discrete Systems So, when the input is, the output will be y[n] H(e )e h[m] e e mn1 j j n -j m j n o o o o where the term in parenthesis in the transient term is basically Fourier transform of h[n] from time n+1 onwards evaluated at ω o. For a FIR impulse response, 0 n N-1, the transient term is zero for n N-1. This implies that the transient response lasts for first N-1 samples only. 41 For IIR stable system, it can be shown that the transient term goes to zero for large n.

Frequency Analysis of LTI Discrete Systems y[n] H(e )e h[m] e e mn1 j j n -j m j n o o o o Total output Steady state x[n] 3 2 1 0-1 42-2 -3-100 -50 0 50 100 n Transient

Frequency Analysis of LTI Discrete Systems Example 4.11: Consider the system with impulse response h[n] = {4, -5, 6, -3} that is excited by x[n] = u[n]. y[n] = 4 x[n] 5 x[n-1] + 6 x[n-2] - 3 x[n-3] n=0 y[0] = 4 x[0] 5 x[-1] + 6 x[-2] - 3 x[-3] = 4 n = 1 y[1] = 4 x[1] 5 x[0] + 6 x[-1] - 3 x[-2] = -1 n=2 y[2] = 4 x[2] 5 x[1] + 6 x[0] - 3 x[-1] = 5 n=3 y[3] =4x[3] 5x[2] +6x[1] -3x[0] = 2 n=4 y[4] = 4 x[4] 5 x[3] + 6 x[2] - 3 x[1] = 2 43

Frequency Analysis of LTI Discrete Systems Example 4.11: Continued. x [ n ] 1 0.8 0.6 0.4 0.2 0 0 5 10 15 20 n y [ n ] 5 4 3 2 1 0-1 0 5 10 15 20 n 44

The Concept of Filtering A digital filter is a system that passes certain frequency components in the input sequence and blocks others. Consider the frequency response 1, 0 j c H(e ) 0, c which represents a lowpass filter Now, if the sequence x[n] = A cos(ω 1 n) + B cos(ω 2 n), with 0<ω 1 <ω c <ω 2 <π, is passed through the system, the output y[n] is j 1 j 2 y[n] A H(e ) cos( n ( )) + B H(e ) cos( n ( )) 1 1 2 2 45 j 1 y[n] A H(e ) cos( n ( )) 0 1 1

The Concept of Filtering Example 4.12: Design a length-3 causal filter h[n] = {α, β, α}that blocks the frequency 0.1 rad/sample and passes the frequency 0.4 rad/sample. Consider the magnitude response only in your design. 46

The Concept of Filtering Example 4.12: Continued. 2 1.5 H(e j ) 1 X: 0.4 Y: 1.001 0.5 47 0 0 01 0.1 02 0.2 03 0.3 04 0.4 05 0.5

The Concept of Filtering Example 4.12: Continued. 3 2 x[n] y[n] cos(0.4n) 1 0-1 -2 48 0 20 40 60 80 100 n

Related functions atan atan2 angle unwrap abs real imag freqz Matlab 49