Unformatted text preview: the unit sequence with the impulse response to get the overall system
response,
n +1
n +1 n +1
1 − − 0.6
1 − 0.6
1 − (−0.8) u[ n ]
n] = 0.6455
0.6455
y[
+
− 1.7746
0.2254
1.8 ( ) 19. Find the impulse responses of these systems:
(a) y′ ( t) + 5 y( t) = x( t)
Follow the example in the text. (b) y′′ ( t) + 6 y′ ( t) + 4 y( t) = x( t)
h′′ ( t ) + 6 h′ ( t ) + 4 h ( t ) = δ ( t )
For t < 0, h( t) = 0 .
For t > 0, h h ( t) = K1e −5.23 t + K 2e −0.76 t
Solutions 316 ( ) M. J. Roberts  8/16/04 Since the highest derivative of “ x ” is two less than the highest derivative of
“y”, the general solution is of the form,
h( t) = (K1e −5.23 t + K 2e −0.76 t ) u( t)
(See the discussion in the text of what the solution form must be for different
derivatives of x and y.) Integrating the differential equation once from t = 0 −
to t = 0 + , [ h′ (0 ) − h′ (0 ) + 6 h(0 ) − h(0
+ − + 0+ − 0+ 0− 0− )] + 4 ∫ h(t)dt = ∫ δ (t)dt = 1 We know that the impulse response cannot contain an impulse because its
second derivative would be a triplet and there is no triplet excitation. We also
know that the impulse response cannot be discontinuous at time, t = 0,
because if it were the second derivative would be a doublet and there is no
doublet excitation. Therefore,
h′ (0 + ) − h′ (0 − ) = 1 ⇒ h′ (0 + ) = 1
This requirement, along with the requirement that the solution be continuous
at time, t = 0, leads to the two equations, [ h′ (0 + ) = 1 = −5.23K1e −5.23 t − 0.76K 2e 0.76 t t = 0+ = −5.23K1 − 0.76K 2 and h(0 + ) = 0 = K1 + K 2 .
(This second equation can also be found by integrating the differential
equation twice from from t = 0 − to t = 0 + .) Solving,
K1 = −0.2237 and K 2 = 0.2237
Then the total impulse response is
h( t) = 0.2237(e −0.76 t − e −5.23 t ) u( t) .
(c) 2 y′ ( t) + 3 y( t) = x′ ( t) (d) 4 y′ ( t) + 9 y( t) = 2 x( t) + x′ ( t) The homogeneous solution is y h ( t) = K h e
h( t) = K h e 3
−t
2 9
−t
4 . The impulse response is of the form, u( t) + K iδ ( t) . Solutions 317 M. J. Roberts  8/16/04 The solution is h( t) = − 9
1 −4t
1
e u( t) + δ ( t)
16
4 20. Sketch g( t) .
(a) g ( t ) = rect ( t ) ∗ rect ( t ) = 1
2 ∞ ∫ rect (τ ) rect (t − τ ) dτ = ∫ rect (t − τ ) dτ −∞ − 1
2 Probably the easiest way to find this solution is graphically through the “flipping and
shifting” process. When the second rectangle is flipped, it looks exactly the same because it
is an even function. This is the “zero shift” position, the t = 0 position. At this position
the two rectangles coincide and the area under the product is one. If t is increased from this
position the two rectangles no longer coincide and the area under the product is reduced
linearly until at t = 1 the area goes to zero. Exactly the same thing happens for decreases in t
until it gets to 1. The convolution is obviously a unit triangle function. This fact is the
reason the unit triangle fu...
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This note was uploaded on 06/19/2013 for the course ENSC 380 taught by Professor Atousa during the Spring '09 term at Simon Fraser.
 Spring '09
 ATOUSA

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