Homework 7 Solution EE35, Spring. Find the Fourier transform of the following signals using tables: (a) te t u(t) h(t) H(jω) te t u(t) ( + jω) (b) sin(πt)e t u(t) h(t) sin(πt)e t u(t) () h(t) ( ejπt e jπt )e t u(t) j h(t) j ejπt e t u(t) j e jπt e t u(t) e t u(t) + jω e jαt s(t) S(j(ω α)) H(jω) j ( + j(ω π) + j(ω + π) ) Or using your tables h(t) sin(πt)e t u(t) () H(jω) π (π) + ( + jω) (c) d (te t sin(t)u(t))
d (te t sin(t)u(t)) d (te t u(t) ejt e jt ) j d j (te t u(t)e jt te t u(t)e jt ) x(t) d j (g(t)) X(jω) j jω(g(jω)) X(jω) ω G(jω) g(t) te t u(t)e jt te t u(t)e jt g(t) h(t)e jt h(t)e jt G(jω) H(j(ω )) H(j(ω + )) h(t) te t u(t) H(jω) ( + jω) G(jω) ( + j(ω )) ( + j(ω + )) X(jω) ω ( ( + j(ω )) ( + j(ω + )) ). Find the inverse Fourier transform of the following signals (Use of tables is acceptable. Requires partial fractions): (a) X(jω) jω ω +5jω+4 X(jω) jω ω + 5jω + 4 A jω + 4 + B jω + Solving for A and B yeilds the result X(jω) jω + 4 jω + x(t) (e 4t e t )u(t)
(b) X(jω) 6jω+6 (jω) +5jω+6 X(jω) Solving for A and B yeilds the result X(jω) 6jω + 6 (jω) + 5jω + 6 A jω + 3 + B jω + jω + 3 + 4 jω + x(t) (e 3t + 4e t )u(t) 3. State in words what the LTI systems do with these transfer functions: (a) H(jω) 5 Amplifies the input by 5. (b) H(jω) Inverts the input. (c) H(jω) e 3jω Delays the input signal by 3. (d) H(jω) e jω Amplifies the input by and delays it by. 4. For an LTI system, y(t) h(t) x(t). (a) Express the system s transfer function H(jω) in terms of the input signal Fourier transform X(jω) and output signal Fourier transform Y (jω). Y (jω) H(jω)X(jω) H(jω) Y (jω) X(jω) (b) Find H(jω) for the LTI system: y(t) x(t) 8x(t 3) y(t) x(t) 8x(t 3) Y (jω) X(jω) 8e 3jω X(jω) Y (jω) X(jω)( 8e 3jω ) H(jω) 8e 3jω 3
(c) Find H(jω) for the LTI system: y(t) y(t 3) 5x(t) y(t) y(t 3) 5x(t) Y (jω) e 3jω Y (jω) 5X(jω) Y (jω)( e 3jω ) 5X(jω) H(jω) 5 e 3jω 5. A causal LTI system is described by the following ordinary differential equation d y(t) + 5 dy(t) + 6y(t) dx(t). (a) Find the frequency response H(jω) for this system. d y(t) + 5 dy(t) + 6y(t) dx(t) (jω) Y (jω) + 5jωY (jω) + 6Y (jω) jωx(jω) H(jω) Y (jω) X(jω) H(jω) jω 6 + 5jω + (jω) 3 3 + jω + + jω (b) Also find its impulse response h(t). h(t) ( 3e 3t + e t )u(t) 6. Use the duality property to find the FT of x(t) +t. First we recognize that By replacing ω with t we get, f(t) e a t F (jω) F (jt) 4 a a + t. a a + ω.
Letting a as in out problem yields Duality states, Therefore our Fourier Transform is X(jt) + t. X(jt) πx( ω). X(jω) πe ω. 7. A signal x(t) cos(t) is the input to an LTI system with frequency response H(jω) 8jω. What is the output signal (in time)? x(t) cost(t) X(jω) πδ(ω ) + πδ(ω + ) π Y (jω) H(jω)X(jω) 8jω [πδ(ω ) + πδ(ω + )] 8j y(t) Working backward from e jt (+8j) + 65 cos(t) 8 65 sin(t) 65 cos(t) 8 65 sin(t) e jt + e jt + 8 e jt e jt 65 65 j e jt 3 + ejt 3 8je jt 3 + 8jejt 3 ( 8j)e jt + ( + 8j)e jt 3 ( 8j)e jt ( + 8j)ejt + 3 ( 8j)( + 8j)e jt 3( + 8j) π δ(ω ) + δ(ω + ) + 8j ejt ( 8j) 65 cos(t) 8 65 sin(t) 3 + 65e jt 3( + 8j) + 65e jt 3( 8j) e jt ( + 8j) + e jt ( 8j) ( + 8j)( 8j)ejt 3( 8j) 5
8. The Fourier transform of a causal signal x(t) is X(jω) (a) Find the total energy of this signal. energy of x(t) + jω. x(t) e t u(t) x(t) e t u(t) e t (b) Find the time t (> ) such that the signal x(t) on the interval [, t ] contains 9% of its total energy. t.9.45 t t x(t) e t u(t) e t e t +. e t ln(.) t t ln(.)/.5 (c) Find the frequency ω (> ) such that the signal x(t) within the frequency band [ ω, ω ] contains 9% of its total energy. By Parseval s theorem, x(t) X(w) dw π energy of x(t) energy of X(w) π 6
so the energy of X(w) is π..9π w w X(w) dw w w w since tan (z) is an odd function + jw dw w + w dw tan (w) w w tan (w ) tan ( w ).9π tan (w ) w tan(.9π ) 6.3 rad/s 9. Consider an LTI system with impulse response h(t) e 4 t. (a) Show the frequency response of this LTI system. H(ω) e (4 jω)t + (b) We have an input e 4t e jωt + e ( 4 jω)t x (t) + n e 4t e jωt 4 jω + 4 + jω 8 ω + 6 δ(t n). Is the output a periodic signal? Justify your answer. If yes, find its Fourier series coefficients. This is a periodic pulse train with period T. The following Fourier Series pair can be applied: n Therefore the output is: 8 π ω + 6 k δ(t nt ) π T δ(ω kπ) 6π 7 k k δ(ω kπ T ) 4k π δ(ω kπ) + 6
So the output in time domain is: k 8 4k π + 6 ejkπt Which is a periodic signal with period. As a result, the Fourier Series Coefficients are (for integer values of k): 8 4k π + 6 Another way to observe the periodicity of the output: The system response to + n δ(t n) is + n h(t n) + n e 4 t n, which is a periodic signal with period. (c) Repeat the same questions for another input x (t) + n ( ) n δ(t n). The input is a series of alternating positive and negative deltas that can be rewritten as: x (t) X(ω) π + n k Therefore the output is: 8π ω + 6 ( e jω ) k δ(t n) So the output in time domain is: + n δ(ω kπ ) e jω π k δ(ω kπ) 8π δ(t (n + )) k 4( ( ) k ) k π + 6 ejkπt k δ(ω kπ ) ( ) k k π δ(ω kπ) + 6 Which is periodic with period. The Fourier series coefficients are: 4( ( ) k ) k π + 6 Another way to observe the periodicity of the output: The system response to + n ( )n δ(t n) is + n ( )n h(t n) + n ( )n e 4 t n, which is a periodic signal with period. 8
. Let y(t) (x(t) cos t) sin t πt. Assuming that x(t) is band-limited and X(jω) for ω, does there exist an LTI system T { } such that T {x(t)} y(t)? Justify your answer. If yes, give the impulse response of this LTI system. From trigonometric identities we have: { y(t) x(t) + cos(t) } sin(t) πt Next we take the Fourier transform: { Y (jω) X(jω) δ(ω) + π δ(ω ) + π } δ(ω + ) rect( ω ) Y (jω) {X(jω) + πx(j(ω )) + πx(j(ω + ))} rect(ω ) Since the signal is band-limited and rect( ω ) function is zero for ω > the X(j(ω + )) and X(j(ω )) terms disappear and we are left with: Y (jω) X(jω) Y (jω) X(jω)H(jω) H(jω) Taking the inverse Fourier transform yields: h(t) δ(t). From previous lectures we know that the energy of a complex signal x(t) is, W x x(t) Noting that x(t) x(t)x (t) determine the relationship between finding the energy of a signal in the time domain and finding the energy of the signal in the frequency domain. This is an important relation called Parsivals Theorem. (Hint: determine what x (t) is equal to in the frequency domain as your first step) Given X(ω) we can write x(t) and x (t) as: x(t) π X(ω)e jωt dω x (t) π X (ω)e jωt dω 9
Next, we can substitute into the following equation replacing one of the ω s with α: Reordering: ( X (jω) π x(t) x(t)x (t) ( ) x(t) X (jω)e jωt dω π x(t)e jωt ) dω X (jω)x(jω)dω π. Use Parsivals Theorem to solve the following problem, Using Parseval s Theorem: χ jω + dω e t u(t) jω + dω 4π jω + e t u(t) 4π e 4t 4π 4 ( ) π