Transcript ee211_8 - University of Kentucky

Pulse Response - First Order
Circuits
Unit Step functions, Pulse Sources,
and SPICE
Kevin D. Donohue, University of Kentucky
1
The Unit Step Function
The unit step function models the behavior of a switch
(i.e. off-on) and can be used to describe more complex
circuit sources.
 0 for t  0
u (t )  
1 for t  0
Multiplying u(t) by a constant changes the value of u(t) for t > 0.
Subtracting a positive number from the argument of u(t) shifts the step in
the positive t direction.
Adding step functions together results in piece-wise steps with the
resultant values equal to the sum of step values in each t interval.
Kevin D. Donohue, University of Kentucky
2
Example
Determine the unit step response for io(t) the circuit
below with input vs(t) = u(t):
io(t)
10 
0.1 mF
vs(t)
40 
Step Response
Show:
18
0.8  0.8 exp( 1250 t ) u (t ) amps
(t ) 
40
16
14
12
ma
iostep
20
10
8
6
4
2
0
Kevin D. Donohue, University of Kentucky
0
1
2
ms
3
4
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The Unit Pulse Function
The unit pulse function is defined as :
1.4
Pulse Function for T = 5
1
0.8
Volts
 0 for t  0

v(t ; T )  1 for 0  t  T
0 for
t T

1.2
0.6
0.4
0.2
0
-10
-5
0
5
10
15
Time
It can be expressed in terms of a combination of unit
step functions:
v(t )  ut   ut  T 
Kevin D. Donohue, University of Kentucky
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The Unit Ramp Function
The unit ramp function is defined as :
0 for
r (t )  
 t for
t0
t0
It can be expressed as the integral of the unit step:
r (t ) 
t
 u( )d

Kevin D. Donohue, University of Kentucky
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The Unit Impulse Function
The unit impulse function is defined as :
 0 for

 (t )   for
 0 for

t0
t 0
t 0
It can be expressed as the derivative of the unit step:
du (t )
 (t ) 
dt
0
Note that:
1    (t )dt
Kevin D. Donohue, University of Kentucky
0
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Example
Determine the pulse
response for io(t) the
circuit below with input
vs(t) = 40v(t) V, where
pulse duration T is 2 ms:
io(t)
10 
vs(t)
0.1 mF
40 
Show:
iopulse(t )  0.8  0.8 exp(1250t )u(t )  0.8  0.8 exp(1250(t  2m))u(t  2m)amps
iopulse(t )  0.8u(t )  u(t  2m)  0.8 exp(1250t )u(t ) 12.1825u(t  2m)amps
Pulse Response
0.8
0.7
0.6
ma
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
ms
5
6
7
8
Kevin D. Donohue, University of Kentucky
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SPICE Simulation with Step Source
Do a SPICE Simulation to determine the pulse response for io(t) the
Previous circuit with input vs(t) = 40v(t) V, where pulse duration T is 2 ms:

To solve you must do a transient analysis, define the voltage source as piece-wise
linear, and describe its transient properties.
R
10
(Amp) +0.000e+000
+2.000m
+4.000m
R0
40
tranex-Transient-14
C
0.0001
+8.000m
V
0
tranex-Transient-14-Table
TIME
I(VAM)
(s)
(Amp)
+0.000e+000
+0.000e+000
+10.000n
+99.999n
+10.840n
+109.104n
:
:
:
:
:
:
+7.285m
+972.534u
+7.445m
+795.710u
+7.605m
+651.035u
+7.765m
+532.665u
+7.925m
+435.817u
VAm
Time (s)
+6.000m
+8.000m
+500.000m
+396.630u
+0.000e+000
I(VAM)
Kevin D. Donohue, University of Kentucky
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SPICE Simulation with Switch
Do a SPICE Simulation to determine the pulse response for io(t) the circuit
below with input vs(t) = 40 V, where pulse duration T is 2 ms:

To solve you must do a transient analysis, define auxiliary circuits for voltage
sources and meters that control the opening and closing of the switches.
SwitchV0
R
10
VA m
R0
40
C
0.0001
SwitchV1
V
40
tranexsw-Transient-0
(Amp)+0.000e+000
+2.000m
+4.000m
Time (s)
+6.000m
+8.000m
+600.000m
(Amp)
+400.000m
+200.000m
V1
0
V0
0
IVm
IVm0
+0.000e+000
I(VAM)
Kevin D. Donohue, University of Kentucky
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