10.1
The Ramp, Step, and Impulse Function:
The following functions will be use to model the behavior of various switch and
source combinations.
The Unit Ramp Function:
0
r (t )  
t
for t  0
for t  0
1. Multiplying r(t) by a constant A changes the slope of r(t) for t > 0.
2. Subtracting a positive number from the argument of r(t) shifts the ramp in the
positive t direction.
3. Adding ramp functions together results in piece-wise ramps with the resultant
slope equal to the sum of slopes in each t interval.
Examples:
10.2
The Unit Step Function:
0
u( t )  
1
for t  0
for t  0
1. Multiplying u(t) by a constant A changes the value of u(t) for t > 0.
2. Subtracting a positive number from the argument of u(t) shifts the step in the
positive t direction.
3. Adding step functions together results in piece-wise steps with the resultant
values equal to the sum of step values in each t interval.
Examples:
10.3
The Unit Impulse Function:
0
d
d

r (t )  u(t )   (t )  
dt
dt
0

2
for t  0
for t  0
for t  0
Important Properties of the Unit Impulse function:
The integral of (t) over any t interval that includes the point where the argument
becomes zeros is one:
t 0
  ( t  t ) dt  1
t 0
0
When the impulse function is present in the integrand with another function, the
resultant integral is simply the function evaluated at the value where the argument
of the delta function goes to zero:
for a  t  b
 f (t )
f
(
t
)

(
t

t
)
dt



elsewhere
 0
Examples:
b
0
0
a
0
10.4
One way to get a better mathematical understanding of the impulse function is to
consider it as a limit of another function:
1

 
f (t )  r  t    r  t   
   2  2 
What is the limit for f(t) as 0?
Note that :
g (t ) 
d
1

 
f (t )  u t    u t   
dt
   2  2 
What is the limit for g(t) as 0?
For any , what is the area under g(t)?
10.5
Singularity Function Examples:
The ramp, step, and delta function are singularity functions. These are functions
that have points where derivatives (or higher order derivatives do not exist).
Functions for which all orders of derivatives exist (i.e. sinusoids, exponentials,
polynomials, ...) are called analytic.
Find and sketch the derivative of:
f (t )  10u(t  3)  6r (t  2)  12u(t )  6r (t  2)
Find and sketch the integral of:
f (t )  0.5u(t  3)  2 (t )  u(t )   (t  1)
Denote the function below in terms of singularity functions:
f(t)
3
2.5
2
volts
1.5
1
0.5
0
-0.5
-1
-2
-1
0
1
seconds
2
3
4
10.6
Response to First-Order Circuits to Singularity Function Sources:
Determine the unit ramp, step and impulse response for the circuit below with input
v ( t ) and output i ( t ) :
io(t)
10 
s
o
40 
0.1 mF
vs(t)
Step Response
Ramp Response
iostep (t ) 
40
0.8  0.8 exp(1.25kt ) u (t )
16
0.05
amps
14
12
0.04
ma
ioramp (t ) 
.8t  0.64m exp(1.25kt )  .64mu (t )
18
0.06
ma
Show:
20
0.07
0.03
10
8
6
0.02
4
0.01
amps
0
40
1000 exp(1.25kt ) u (t )
iostep (t ) 
amps
40
2
0
0
1
4
2.5
2
ms
3
4
Impulse Response
x 10
2
ma
1.5
1
0.5
0
0
1
2
ms
3
4
0
1
2
ms
3
4
10.7
The Ramp Response:
The ramp response represents a new type a source whose waveform is simply a
linear ramp. The fact that ramp is a singularity function implies the switch (used in
previous problems) is now modeled in the source. The new feature in finding the
complete response to a ramp input is the way in which the force response is
determined. All other aspects of the problem remain the same.
Example: Find the complete response for vL(t) in the circuit below where the source
vs(t) is 800r(t):
400 
vs(t)
0.01 H
+
vL(t)
-
100 
Show that inductor current iL(t) and vL(t) are given by:
 1

i (t )  
exp( 8000t )  1  2t  u(t ) and v (t )  0.021  exp( 8000t )u(t )

 4000

L
L
10.8
The Step Response:
The step response represents a DC source with a switch. The new feature in finding
the complete response to a step input is simply the notation.
Example: Find the complete response for vL(t) in the circuit below where the source
vs(t) is 800u(t):
400 
vs(t)
0.01 H
+
vL(t)
-
100 
Show that inductor current iL(t) and vL(t) are given by:
i (t )  21  exp( 8000t )u(t ) and v (t )  160 exp( 8000t )u(t )
L
L
10.9
The Impulse Response:
The impulse response represents a new type of source. Although it is not a practical
source, the impulse response provides a characterization of the circuit analogous to
the transfer function. In practice, applying a large voltage or current for a very
short duration of time (a spike) can approximate this source. The forced response
to an impulse is 0, since for t > 0, (t) = 0. The impulse source effectively injects
energy into the circuit and lets the circuit release the energy according to its natural
response.
Example: Find the complete response for vL(t) in the circuit below where the source
vs(t) is 800(t):
400 
vs(t)
0.01 H
+
vL(t)
-
100 
Show that inductor current iL(t) and vL(t) are given by:
i (t )  16000 exp( 8000t )u(t ) and v (t )  128
. M exp( 8000t )u(t )  160 (t )
L
L