MCTE 2311
Signal and System Analysis
Lecturer:
Azhar Bin Mohd Ibrahim (Ph.D. Universiti Sains Malaysia)
azhar_ibrahim@iium.edu.my
(Room: E1-4-2.9)
Basic Signals and Systems
Contents
• Basic building blocks of signals
• Types of systems
Useful Signals
Unit Step Function, u(t)/ Sequence u(n)
u(t)
1
0
t
u(n)
1
0
n
Rectangular Pulse
Consider the rectangular pulse x(t) with an amplitude A and duration
of 1 second. Express x(t) as a weighted sum of two step functions.
Rectangular pulse x(t):
Rectangular Pulse
Representation of x(t) as the
difference of two step functions
of amplitude A, with one step
function shifted to the left by ½
and the other shifted to the
right by ½; the two shifted
signals are denoted by x2(t) and
x1(t), respectively.
So, x(t) = x2(t) – x1(t).
Ramp Function
Continuous-time:
Importance of unit impulse
Because any continuous signal is a linear combination of an infinite number of scaled and
time-shifted impulse signals.
Exercise
Sketch:
2u(t-5)
2u(t-9)
-4u(t-7)
Exercise
Sketch:
r(t+1)
-r(t)
Exercise
Sketch:
Exercise
• Express the following triangular pulse in terms
of unit ramp functions.
Exercise
• Express the following signal in terms of unit step and unit ramp
signals
Exercise
Sketch:
Exercise
Sketch:
Exercise
Sketch:
Exercise
Sketch:
Summary of Basic signals
These signals serve as building blocks to other complex signals.
1. Step signal
2. Impulse signal
3. Ramp signal
Time-invariance
Time invariant
System
Time invariant
System
Input delayed by m seconds
Output also delayed by m seconds
Linearity
Time invariant
System
Time invariant
System
Input scaled by m
Output also scaled by m
Linear Systems
• A system is linear when superposition principle
holds for the system.
• Otherwise, a system is said nonlinear.
• Superposition principle:
Linear System
The linearity property of a system.
(a)
The combined operation of amplitude scaling and summation
precedes the operator H for multiple inputs.
(b)
The operator H precedes amplitude scaling for each input; the
resulting outputs are summed to produce the overall output y(t).
If these two configurations produce the same output y(t), the
operator H is linear.
Linear System: Example
• Determine whether the system defined by the
equation y(t) = 2x(t) is linear or not.
The system is linear.
Linear System: Example
• Determine whether the system defined by the
equation y(t) = 2x(t) + 1 is linear or not.
The system is nonlinear.
Linear System: Example
•
The system is non-linear.
Linear System: Example
• Given
.
Verify the linearity of this system.
The system is linear.
Linear System: Example
• Given
.
Verify the linearity of this system.
The system is nonlinear.
Time-Invariant System
• A system is time-invariant if its input-output
relationship does not change with time. That
is, a delayed input produces an equally
delayed output.
• Otherwise it is said to be time-varying.
System
Time-Invariant System: Example 1
• Determine if the system represented by the
equation y(t)= 2x(t)+1 is time-invariant or
time-varying.
• Solution
Delayed input implies x(t – to):
y1(t) = 2 x(t – to) + 1
Delayed output implies y(t – to):
y2(t) = y(t – to) = 2 x(t – to) + 1
Comparing the above results:
y2(t) = y1(t)
The system is time-invariant.
Time-Invariant System: Example 1
• Determine if the system represented by the
equation y(t) = x(t) sin(ω0 t) is time-invariant
or time-varying.
• Solution
Delayed input implies x(t – to):
y1(t) = x(t – to) sin( ω0t )
Delayed output implies y(t – to):
y2(t) = y(t – to) = x(t – to) sin( ω0 (t – to) )
Comparing the above results:
The system is time-varying.
Example 2
• Check the system represented by the
following equation for both linearity and
time-variance: y(t) = t x(t)
• Check for linearity:
The system is linear.
Example 2 (cont.)
• Check for time-variance:
Delayed input:
Delayed output:
The system is time-varying.
System responses
System
Step response
System
Impulse response
Example 1
• If the impulse response of a LTI system is
1
System
0
Impulse response
Unit impulse
• Predict the response of the system for the following input
2
1.5
1
0.5
System
-2
0
2
?
Example 1
• If the impulse response of a LTI system is
1
System
0
Impulse response, h(t)
Unit impulse, δ(t)
• The response of the system for the following input is….
6
2
1.5
3
1
0.5
1.5
System
-2
0
2
-3
x(t) = 0.5 δ(t+2) + 2 δ(t) + 1 δ(t-2)
-2
-1
0
1
2
y(t) = 0.5 h(t+2) + 2 h(t) + 1 h(t-2)
3
Example 2
• If the step response of a LTI system is
1
System
0
step response
Unit step signal
• Predict the response of the system for the following input
2
1.5
1
0.5
System
-2
0
2
?
Example 2
• If the step response of a LTI system is
1
System
0
step response
Unit step signal
• The response of the system for the following input is
2
1.5
3
1
0.5
System
-2
0
2
0
-1
1
2
3
x(t) = u(t) – u(t-2)
-3
y(t) = step response – step response (t – 2)
Example 3: (Midterm question Sem 1 2013-2014)
Pulse signal
Pulse response
Example 3
Characteristic of LTI system
If the input to a LTI system is a sinusoidal signal, the output is also a
sinusoidal signal with the same frequency as the input signal.
x(t)
y(t)
LTI system
1kΩ
+
x(t)
+
1µF
y(t)
Why impulse response is important?
• Because if a system is LTI (linear and time-invariant), we can use
its impulse response to predict its output for any input signal!
• Remember that any signal can be decomposed into a linear
combination of impulses.
• So if the system is LTI, the output will also be a linear
combination of impulse responses!
Causal System
• A system is causal (or non-anticipative) if its output depends on
past/current inputs but not future inputs.
• i.e. the output y(t0) only depends on the input x(t) for values of t < t0.
• Suppose h(t) is the impulse response of the system H.
(only fully accurate for a system described by linear
constant coefficient differential equation)
h(t) = 0 for all t <0
then the system H is causal, otherwise it is
non-causal.
Causal System: Example
Determine whether the systems presented
below are causal or non-causal:
1. y(t) = t x(t)
2. y(t) = (t – to) x(t – to)
3. y(t) = (t + to) x(t + to)
Answer:
1. Causal
2. Causal
3. Non-causal; it consists of future values of
x(t).
Memory System
• A system is instantaneous or memoryless or zero memory if its
present output depends on its input at the present time only.
• A system is dynamic or has memory if its present output depends
on its input at past or future in addition to the present time.
• For example, an inductor is a system with memory because its
current i(t) depends on past values of voltage v(t), as shown
below:
Memory System: Example
Memoryless
• Resistor:
• Moving-average system:
Memory
• A system described by the input-output relation:
Memoryless
Stable System
• A system is said to be bounded-input,
bounded-output (BIBO) stable if and only if
every bounded input results in a bounded
output.
• The operator H is BIBO stable if the output
signal y(t) satisfies the condition
whenever the input signals x(t) satisfy the
condition
Stable System: Example
• Show that the previous moving-average
system is BIBO stable.
Solution
Assume:
System:
Moving
average
system is
BIBO stable
One Famous Example of an
Unstable System
Dramatic photographs showing the collapse of
the Tacoma Narrows suspension bridge on
November 7, 1940.
(a)
Photograph showing the twisting motion of
the bridge’s center span just before failure.
(b)
A few minutes after the first piece of concrete
fell, this second photograph shows a 600-ft
section of the bridge breaking out of the
suspension span and turning upside down as
it crashed in Puget Sound, Washington.
(Courtesy of the Smithsonian Institution)
See also:
http://www.britannica.com/EBchecked/media/411/Collapse-of-the-Tacoma-Narro
ws-Bridge-Washington-state-1940
Inverse of Systems
• A system is said to be invertible if the input of
the system can be recovered from the output.
x(t) = input
y(t) = output
H = first system operator
H inv = second system operator
The second operator Hinv is the inverse
of the first operator H.
Condition for invertible
system:
Inverse of Systems: Example 2
• Determine if a square-law system
described by the input-output relation
invertible or not.
Solution:
Since distinct inputs x(t) and −x(t) produce the
same output y(t), so the square-law system is not
invertible.