hw07-s06-sol-redacted

# hw07-s06-sol-redacted - EECS 20N Structure and...

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EECS 20N: Structure and Interpretation of Signals and Systems Department of Electrical Engineering and Computer Sciences U NIVERSITY OF C ALIFORNIA B ERKELEY Problem Set 7 SOLUTIONS HW 7.1 (a) x ( t ) = e - at u ( t ) , a > 0 X ( ω ) = Z -∞ e - at u ( t ) e - jωt dt = Z 0 e - at e - jωt dt = e - ( a + ) t - ( a + ) | 0 = 0 - 1 - ( a + ) = 1 a + (b) x ( t ) = e - a | t | X ( ω ) = Z -∞ x ( t ) e - jωt dt = Z 0 -∞ e at e - jωt dt + Z 0 e - at e - jωt dt = Z 0 -∞ e ( a - ) t dt + Z 0 e - ( a + ) t dt = e ( a - ) t a - | 0 -∞ + e - ( a + ) t - ( a + ) | 0 = 1 - 0 a - + 0 - 1 - ( a + ) = 1 a + + 1 a - = 2 a a 2 + ω 2 1

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(c) x ( t ) = sgn ( t ) dx ( t ) dt = 2 δ ( t ) y ( t ) = 2 δ ( t ) Y ( ω ) = Z -∞ 2 δ ( t ) e - jωt dt = 2 CTFT dx ( t ) dt = 2 differential property: y ( t ) = dx ( t ) dt = d dt 1 2 π Z -∞ X ( ω ) e jωt = 1 2 π Z -∞ X ( ω ) jωe jωt Y ( ω ) = X ( ω ) For ω 6 = 0 X ( ω ) = Y ( ω ) For ω = 0 X (0) = Z -∞ sgn ( t ) e - jωt dt | ω =0 = Z -∞ sgn ( t ) dt = 0 X ( ω ) = 2 ω 6 = 0 0 ω = 0 (d) x ( t ) = u ( t ) = sgn ( t ) + 1 2 Let y ( t ) = 1 2 Then, 1 2 = 1 2 π Z -∞ Y ( ω ) e jωt dt Y ( ω ) = πδ ( ω ) Using 1(c), X ( ω ) = 1 ω 6 = 0 πδ ( ω ) ω = 0 2
(e) x ( t ) = X k = -∞ δ ( t - kT ) x ( t ) = X k = -∞ X k e ikω 0 t , ω 0 = 2 π T CTFS Expansion X k = 1 T Z T 2 - T 2 δ ( t ) e - ikω 0 t dt = 1 T x ( t ) = 1 T X k = -∞ e ikω 0 t X ( ω ) = 2 π T X k = -∞ δ ( ω - k 2 π T ) HW 7.2 (a) ˆ x ( t ) = x ( at ) ˆ X ( ω ) = Z -∞ x ( at ) e - iωt dt Let t 0 = at, dt 0 = adt if a > 0 t 0 =

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