Signals and Systems Spring 2004 Lecture #9

Transcription

1 Signals and Systems Spring 2004 Lecture #9 (3/4/04). The convolution Property of the CTFT 2. Frequency Response and LTI Systems Revisited 3. Multiplication Property and Parseval s Relation 4. The DT Fourier Transform Figures and images used in these lecture notes by permission, copyright 997 by Alan V. Oppenheim and Alan S. Willsky The CT Fourier Transform Pair x(t) X( jω ) X( jω ) = x(t)e jωt dt FT x(t) = X( jω)e jωt dω Inverse FT Last lecture: Today: some properties further exploration 2

2 Convolution Property y(t) = h(t) x(t) Y( jω ) = H( jω )X( jω) where h(t) H( jω ) Basically a consequence of the eigenfunction property x(t) = X( jω )dω coefficient e jωt a e jωt h(t) H( jω)a e jωt y(t) = = superposition H(jω) X( jω)dω New coefficient 3 e jωt H(jω)X( jω ) ejωt dω Y( jω) The Frequency Response Revisited x(t) h(t) y(t) = h(t) x(t) Y(jω) = H(jω)X( jω ) The frequency response H(jω) of a CT LTI system is simply the Fourier transform of its impulse response h(t) Example #: Recall x(t) = e jω o t H(jω) y(t) e jω o t δ(ω ω o ) Y(jω) = H(jω)X( jω ) = H( jω )δ(ω ω 0 ) = H( jω 0 )δ(ω ω 0 ) inverse FT y(t) = H( jω 0 )e jω 0 t Now it is clear to see frequency shaping effects of earlier demos. 4

3 Example #2: A differentiator y(t) = dx(t) dt From differentiation property d dt an LTI system FT jω H(jω) = jω ) Amplifies high frequencies (enhances sharp edges) 2) +π/2 phase shift (j = e j!/2 ) Larger at high ω o phase shift d dt sinω 0t = ω 0 cosω 0 t = ω 0 sin(ω 0 t + π 2 ) d dt cosω 0 t = ω 0 sinω 0 t = ω 0 cos(ω 0 t + π 2 ) 5 Example #3: Impulse Response of an Ideal Lowpass Filter h(t) = Questions: ) Is this a causal system? No. 2) What is h(0)? h(0) = H(jω)dω = ω c / π 3) What is the steady-state value of the step response, i. e. s()? s() = h(t)dt = H( j0) = ω c ω c 6 e jωt dω = sinω c t πt = ω c π sinc ω t c π sin c(θ) = sin(πθ) πθ

4 Example #4: Cascading filtering operations 7 Example #5: sin 4πt sin8πt =? 2 πt3 πt 23 x(t) h(t) b Y(jω) = X(jω) y(t) = x(t) Example #6: e at 2 e bt 2 =? π a e b Gaussian Gaussian = Gaussian, ω 2 4 a π ω2 4b e = π ω 2 b ab e 4 a + b Gaussian Gaussian = Gaussian 8

5 Review from the last lecture, right-sided exponential x(t) = e at u(t), a > 0 X( jω ) = x(t)e jωt dt = e 4 at 2 e 43 jωt dt 0 e (a +jω)t = e (a + jω )t = a + jω 0 a + jω 9 Example #7: h(t) = e t u(t), x(t ) = e 2t u(t) y(t) = h(t) x(t) =? Y( jω ) = H( jω )X(jω) = (+ jω ) (2 + jω) a rational function of jω, ratio of polynomials of jω Y( jω) = Partial fraction expansion + jω 2 + jω 0 inverse FT y(t) = [e t e 2t ]u(t)

6 Example #8: LTI Systems Described by LCCDE s (Linear-constant-coefficient differential equations) N N d k M y(t) d k x(t) = b dt k k dt k k = 0 d k x(t) dt k k = 0 Using the Differentiation Property ( jω ) k X( jω ) Transform both sides of the equation ( jω )k Y( jω) = b k ( jω )k X(jω) k =0 Y(jω) = M k = 0 M b k ( jω ) k k =0 N X( jω ) (jω) k k = H( jω) ) Rational, can use PFE to get h(t) 2) If X(jω) is rational i.e. x(t) = α i e -βit u(t), then Y(jω) is also rational Parseval s Relation x(t) 2 dt = X( jω ) 2 dω Total energy in the time-domain Since FT is highly symmetric, x(t) = F {X( jω)} = thus if then the other way around is also true Total energy in the frequency-domain Multiplication Property x(t) y(t) X( jω) Y( jω ) x(t) y(t) X( jω ) Y(jω) = X(jθ)Y(j( ω θ))dθ A consequence of Duality 2 2 X( jω) spectral density X( jω) e jωt dω, X( jω) = F{x(t)} = x(t) e jωt dt Definition of convolution in ω

7 Examples of the Multiplication Property r(t) = s(t) p(t) R(jω) = [ S(jω) P(jω) ] For p(t) = cosω 0 t P( jω) = πδ(ω ω 0 ) + πδ (ω + ω 0 ) R( jω ) = 2 S( j(ω ω 0)) + 2 S( j(ω +ω 0)) 3 Example (continued) r(t) = s(t) cos(ω o t) Amplitude modulation (AM) R( jω ) = [S( j(ω ω o )) 2 + S( j(ω +ω o ))] Drawn assume ω o - ω >0 i.e. ω o > ω 4

8 The Discrete-time Fourier Transform Derivation: (Analogous to CTFT except e jωn = e j(ω+)n ) x[n] - aperiodic and (for simplicity) of finite duration N is large enough so that x[n] = 0 if n N/2 x [n] = x[n] for n N/2 and periodic with period N x [n] = x[n] for any n as N 5 DTFT Derivation (Continued) x [n] = k =<N> e jkω 0 n, ω 0 = N = N = N n = N N 2 n= N x [n]e jkω 0 n x [n]e jkω 0 n = N n = x[n]e jkω 0 n Define X(e jω ) = n= x[n]e jωn = N X(e jkω 0 ) periodic in ω with period 6

9 x [n] = DTFT Derivation (Home Stretch) N X(e jkω 0 ) e jkω 0 n = k =<N> As N : k =<N > x [n] x[n] for every n X(e jkω 0 )e jkω 0 n ω 0 ( ) ω o 0, ω dω o x[n] = X(e jω ) = The sum in ( ) an integral X(e jω )e jωn dω Any interval in ω Synthesis equation x[n] e jωn Analysis equation n= The DTFT Pair 7