Post on 14-Dec-2020
SIGNALS AND SYSTEMS: PAPER 3C1
HANDOUT 2.
Dr Anil Kokaram
Electronic and Electrical Engineering Dept.
anil.kokaram@tcd.ie www.mee.tcd.ie∼ack
SYSTEM RESPONSE
• Ultimate goal is to control a process or analyze a signal To do this need
to be able to predict the response of a system to various inputs
• Given known system dynamics (i.e. we know enough about the system
to write down ODE’s describing its operation) we can use the Laplace
Transform to work out the response of the system to any input whose
Laplace Transform can be found.
• However, real inputs are not predictable; and it may not be possible to
write down a system’s ODE’s. The process used in Mobile Phones for
canceling echoes is a good example of a system which operates despite
the inability to analytically define the geometry/dynamics of the phone’s
immediate environment.
• Still can identify some basic system characteristics which allow handling
of system response. This is possible using purely time domain analysis.
• Time domain analysis relies on the fact that the response of a system to an
impulse or to a step tells you everything about the system (in principle).
These responses can be calculated if the ODE’s for the system can be
written, or they can be measured by experiment beforehand or on-line
(e.g. mobile phones, noise cancellation etc.)
• We will explore both techniques for analyzing systems. Start with Laplace
analysis as you met this in 2nd year.
3C1 Signals and Systems 1 www.mee.tcd.ie/∼sigmedia
1 REVIEW OF LAPLACE ANALYSIS
1 Review of Laplace Analysis
We’ll be using the Laplace Transform to solve differential equations.
Need to be confident in using it.
1. Laplace Transform of a signal f (t) is
L(
f (t)
)=
∫ ∞
0
f (t)e−stdt (1)
s is a COMPLEX number, e.g. σ + jω !! The result of taking
the Laplace Transform of a signal f (t) is a function in s. So we
go from a 1-dimensional function i.e. a function of time only, to
a function of 2 variables σ and ω i.e. a function of a complex
variable. We will write the laplace Transform of f (t) as F(s).
Sometimes in books you might see f (s) whic is useful because
some capital letters look like the common ones when written by
hand e.g. V and v.
2. Inverse laplace Transform of F(s) is
f (t) = L−1
(F(s)
)=
1
2πj
∫ σ+j∞
σ−j∞F(s)estds (2)
This is a nasty1 integral to do, need to know about Contour
integration. Happily, being Engineers we can use tables of Laplace
Transform pairs to do this.
3. Sometimes people panic about the convergence of the inverse
Laplace Transform, because you find yourself having to say, for
instance ∫ ∞
0
e−stdt =
[−e−st
s
]∞
0
=1
s(3)
1Not really, but it makes you feel better if you think this is so.
3C1 Signals and Systems 2 www.mee.tcd.ie/∼sigmedia
1 REVIEW OF LAPLACE ANALYSIS
even though you don’t know what value s is. Don’t panic, in this
course we are going to assume that the Laplace integral always
converges (i.e. Re(s) > 0) and so[−e−st
s
]∣∣∣∣s=∞
= 0 (4)
BUT YOU NEED TO REMEMBER THAT CONVERGENCE
IS AN ISSUE
4. In addition, assume that e−(s+a)t is 0 for s = ∞, (i.e. you can
always assume that s + a > 0 to make this happen.)
5. Remember to spot the ‘shift’ theorem since it gives a slick way of
using tables to get Laplace transforms of functions (we’ll derive
this later for the Fourier transform as well).
Given L(
f (t)
)= F(s)
Then L(
e−ktf (t)
)= F(s + k)
In other words, if you want to find the Laplace transform
of e−ktf (t), then just find the Laplace transform of f (t)
and replace all occurrences of s with s + k. A similar
statement can be made for the inverse transform.
3C1 Signals and Systems 3 www.mee.tcd.ie/∼sigmedia
1.1 Transform Tables 1 REVIEW OF LAPLACE ANALYSIS
1.1 TABLE OF LAPLACE TRANSFORM RELATIONS
Waveform: Laplace Transform:
g(t) (defined for t ≥ 0) G(s) = Lg(t) =∫∞
0−g(t)e−stdt
δ(t) impulse 1
u(t) unit step 1s
tn n!sn+1
e−at 1s+a
sin(ω0t)ω0
s2+ω20
cos(ωot)s
s2+ω20
sinh(ω0t)ω0
s2−ω20
cosh(ω0t)s
s2−ω20
e−at[A cos(ω0t) + B sin(ω0t)]A(s+a)+Bω0
(s+a)2+ω20
e−atg(t) G(s + a) shift in s
g(t− τ)u(t− τ) where τ ≥ 0 e−sτG(s) shift in t
tg(t) − ddsG(s)
dgdt differentiation sG(s)− g(0)
d2gdt2 2nd differential s2G(s)− sg(0)−
(dgdt
)∣∣∣∣0
dngdtn snG(s)− sn−1g(0)− sn−2
(dgdt
)∣∣∣∣0− . . .−
(dn−1gdtt−1
)∣∣∣∣0∫ t
0 g(τ)dτ integration G(s)s
g1(t) ∗ g2(t) convolution G1(s)G2(s)
=∫ t
0 g1(t− τ)g2(τ)dτ
3C1 Signals and Systems 4 www.mee.tcd.ie/∼sigmedia
1.2 Laplace Transform Examples 1 REVIEW OF LAPLACE ANALYSIS
1.2 Laplace Transform Examples
1. Find the Laplace Transform of e−at from first principles
Le−at =
∫ ∞
0
e−ate−stdt
=
∫ ∞
0
e−(s+a)tdt
=
[− 1
s + ae−(s+a)t
]∞
0
= − 1
s + a
[0− 1
]
=1
s + a
2. Find the Laplace Transform of δ(t− a) from first principles
Lδ(t) =
∫ ∞
0
δ(t)e−stdt
= 1
3. Find the Laplace Transform of δ(t− a) from first principles
Lδ(t− a) =
∫ ∞
0
δ(t− a)e−stdt
= e−at
4. Laplace transform of e−at + e−bt. Laplace transform obeys super-
position so..
Le−at + e−bt = Le−at + Le−bt=
1
s + a+
1
s + b
3C1 Signals and Systems 5 www.mee.tcd.ie/∼sigmedia
1.2 Laplace Transform Examples 1 REVIEW OF LAPLACE ANALYSIS
5. Find the Laplace transform of e−at sin(βt + φ)from first principles (i.e. the hard way) :
L
e−at sin(βt + φ)
=
∫ ∞
0e−at sin(βt + φ)e−stdt
Using sin(x) =1
2j
(ejx − e−jx
)
=
∫ ∞
0e−at 1
2j
(ej(βt+φ) − e−j(βt+φ)
)e−stdt
=1
2j
∫ ∞
0e−(s+a)t
(ej(βt+φ) − e−j(βt+φ)
)dt
=1
2j
∫ ∞
0
(ejφe−(s+a−jβ)t − e−jφe−(s+a+jβ)t
)dt
=1
2j
(ejφ
[e−(s+a−jβ)t
−(s + a− jβ)
]∞
0− e−jφ
[e−(s+a+jβ)t
−(s + a + jβ)
]∞
0
)
=1
2j
(ejφ
(s + a− jβ)− e−jφ
(s + a + jβ)
)
=1
2j
((s + a + jβ)ejφ − (s + a− jβ)e−jφ
(s + a)2 + β2
)
=1
2j
((s + a)(ejφ − e−jφ) + jβ(ejφ − e−jφ)
(s + a)2 + β2
)
=(s + a) sin(φ)
(s + a)2 + β2 +β cos(φ)
(s + a)2 + β2
3C1 Signals and Systems 6 www.mee.tcd.ie/∼sigmedia
1.2 Laplace Transform Examples 1 REVIEW OF LAPLACE ANALYSIS
But using tables,
sin(ω0t) ↔ ω0
s2 + ω20
cos(ω0t) ↔ s
s2 + ω20
e−at sin(ω0t) ↔ Using the shift theorem: Replace s with s + a
↔ ω0
(s + a)2 + ω20
we can do the same thing the slick way (i.e. from tables and
spotting the shift theorem) :
L
e−at sin(βt + φ)
=
= L
e−at[sin(βt) cos(φ) + cos(βt) sin(φ)]
= L
e−at sin(βt) cos(φ)
+ L
e−atcos(βt) sin(φ)
= cos(φ)L
e−at sin(βt)︸ ︷︷ ︸
+ sin(φ)L
e−at cos(βt)︸ ︷︷ ︸
=β cos(φ)
(s + a)2 + β2+
(s + a) sin(φ)
(s + a)2 + β2
3C1 Signals and Systems 7 www.mee.tcd.ie/∼sigmedia
1.3 Partial Fractions 1 REVIEW OF LAPLACE ANALYSIS
1.3 Review of partial fractions
You will need to find the inverse Laplace Transform of sometimes complicates
expressions. The method of partial fractions allows you to do this by express-
ing a complicated fraction in terms of a sum of simpler fractions. The idea
is that the inverse laplace transform of the simpler fractions is easier to spot.
Basic ideak3s + 1
s(s + k1)(s + k2)=
A
s+
B
s + k1+
C
s + k2(1)
Expand r.h.s. and equate coeffs
=A(s + k1)(s + k2) + Bs(s + k2) + Cs(s + k1)
s(s + k1)(s + k2)
Equating coeffs in s2 ⇒0 = A + B + C
Equating coeffs in s ⇒k3 = A(k1 + k2) + Bk2 + Ck1
Equating constants ⇒1 = Ak1k2
Then solve simultaneous equations for A, B, C. Urrgh, could be a
pain. There’s another, easier way (cover up rule)
Multiply eqn. 1 by s, set s = 0, This gives A straightaway
⇒ k3s + 1
(s + k1)(s + k2)=
As
s+
Bs
s + k1+
Cs
s + k2(2)
s = 0 ⇒ 1
k1k2= A
Rule is Set s to be value that makes a denominator factor equal
0. Then COVER UP that factor in denominator and subsitute
to get coefficient value. To get C then . . .
s = −k2 ⇒ −k3k2 + 1
−k2(k1 − k2)= C (3)
3C1 Signals and Systems 8 www.mee.tcd.ie/∼sigmedia
2 SYSTEM RESPONSE (AT LAST)
2 System Response (at last)
• What is the response of the system shown to a unit impulse input
at x(t), i.e. What happens to y(t) when x(t) = δ(t) Volts ?
C y(t)x(t)
R
• Steps are
1. Write down the differential equations for modeling the system.
2. Given the stated initial conditions (if any) get an expression for y(t)
in terms of x(t). In other words : solve the differential equations.
3. Do this by
(a) First principles using P.I. and Homogeneous solution. (A pain).
(b) Laplace Transforms (much easier).
4. Insert the numerical values or expression for the input x(t) as well as
initial conditions.
5. Solution falls out.
3C1 Signals and Systems 9 www.mee.tcd.ie/∼sigmedia
2.1 Example 1 : 2 SYSTEM RESPONSE (AT LAST)
2.1 Example 1 :
Impulse response of this system is the response y(t) when x(t) = δ(t).
C y(t)x(t)
R
Write down differential equations to model the system.
x(t) = Ri(t) + y(t) = RCdy
dt+ y(t)
Take Laplace Transforms, and solve for Y(s)
X(s) = RC(sY(s)− y(0)) + Y(s)
Y(s)
(1 + RCs
)= X(s)−RCy(0)
⇒ Y(s) =X(s)−RCy(0)
1 + RCs(4)
Assuming initially, output is zero i.e. for t < 0, y(t) = 0
=1
1 + RCsX(s) = G(s)X(s) (5)
Find Laplace Transform of input x(t): X(s) = Lδ(t) = 1
Hence the impulse response, (we’ll refer to it as h(t) ) is therefore
h(t) = L−1
[G(s)X(s)
]= L−1
[G(s)× 1
]
= L−1
[1
1 + RCs
]= L−1
[1/(RCs)
(1/(RCs) + s
]
= (6)
3C1 Signals and Systems 10 www.mee.tcd.ie/∼sigmedia
2.2 Step response 2 SYSTEM RESPONSE (AT LAST)
2.2 Step response
Impulse response of this system is the response y(t) when x(t) = u(t)
(i.e. x(t) is a step function.
C y(t)x(t)
R
⇒ Y(s) =1
1 + RCsX(s) (7)
Find Laplace Transform of input x(t) : Lu(t) = 1s
Hence the step response y(t) is
y(t) = L−1
[Y(s)
]
= L−1
[1
1 + RCsX(s)
]
= L−1
[(1
1 + RCs
)(1
s
)]
use partial fractions:1
s(1 + RCs)=
A
s+
B
1 + RCs
Use Cover up rule to find coefficients A,B
Hence y(t) = L−1
[1
s− RC
1 + RCs
]
=
= u(t)− e−t/RC
=
3C1 Signals and Systems 11 www.mee.tcd.ie/∼sigmedia
2.3 PLOTS 2 SYSTEM RESPONSE (AT LAST)
2.3 PLOTS OF THE STEP AND IMPULSE RESPONSE
(assuming zero initial conditions, R=10KΩ, C = 10µF. )
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (secs)
Am
plitu
de (
volts
)
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (secs)
Am
plitu
de (
volts
)
3C1 Signals and Systems 12 www.mee.tcd.ie/∼sigmedia
3 IMPULSE AND STEP RESPONSE
3 A time domain relationship
• The step function is the integral of the delta function.
u(t) =
∫ t
0
δ(τ )dτ (8)
• All systems we deal with are LTI so SUPERPOSITION APPLIES
So if input xi gives output yi then∑
i xi gives output®
©
ª.
• Hence
Input
∫ t
0
x(τ )dτ → Output
∫ t
0
y(τ )dτ (9)
• So input u(t) (which is the integral of δ(t)) gives output
• Therefore The step response of a system is the integral of the
impulse response.
• Now do the example 1 again, but this time integrate the impulse
response to get the step response.
3C1 Signals and Systems 13 www.mee.tcd.ie/∼sigmedia
4 TRANSFER FUNCTIONS
4 TRANSFER FUNCTIONS
Recall for the capacitor and resistor circuit we had
Y(s) =X(s)−RCy(0)
1 + RCs
=X(s)
1 + RCs− RCy(0)
1 + RCs
⇒ Y(s) = F (s)X(s) + G(s)y(0) (10)
F (s) is called the transfer function of the system. Its all we need
to work out everything about the system if the initial conditions were
0. The Transfer function generalizes the idea of ‘Gain’ to dynamic
attributes of the system.
+
LAPLACE TRANSFORM OF OUTPUT SIGNAL
LAPLACE TRANSFORM OF INPUT SIGNAL
SYSTEM TRANSFER FUNCTION
OTHER TERMS
(INITIAL CONDITIONS)
= X
So the system block diagram can be drawn as follows.
3C1 Signals and Systems 14 www.mee.tcd.ie/∼sigmedia
5 LINKING H(T ) AND H(S)
5 A link between Impulse response
and Transfer Function
Suppose Impulse response of system is h(t). What is its Transfer
Function?
3C1 Signals and Systems 15 www.mee.tcd.ie/∼sigmedia
6 TRANSFER FUNCTION EXAMPLES
6 Transfer Function examples
Transfer Function G(s) =Laplace Transform of Output Signal
Laplace Transform of Input signal
=Y(s)
X(s)
y(t)RCx(t)
A
+x(t) y(t)
R
R
1
2
-
C
+x(t) y(t)
RA
-
3C1 Signals and Systems 16 www.mee.tcd.ie/∼sigmedia
6.1 A cascade of electrical systems 6 TRANSFER FUNCTION EXAMPLES
6.1 A cascade of electrical systems
+w(t) z(t)
x(t) y(t)
+
--
Assuming that each stage does not load the preceeding one (i.e. each
input impedance >> output impedance of previous stage), Then
X(s)
w(s)= −k1;
Y(s)
X(s)=
1
1 + sT;
z(s)
Y(s)= −k2
3C1 Signals and Systems 17 www.mee.tcd.ie/∼sigmedia
7 BLOCK DIAGRAM ALGEBRA
7 Block Diagram Algebra
3C1 Signals and Systems 18 www.mee.tcd.ie/∼sigmedia
8 MORE BLOCK DIAGRAM ALGEBRA
8 More Block Diagram Algebra
3C1 Signals and Systems 19 www.mee.tcd.ie/∼sigmedia
8 MORE BLOCK DIAGRAM ALGEBRA
EXAMPLES OF THE DISTINCTION BETWEEN
TRANSFORMS OF SIGNALS AND TRANSFER
FUNCTIONS OF SYSTEMS.
SIGNALS SYSTEMS
1 ↔ δ(t) Y(s) = 1 ·X(s) ↔ y(t) = x(t)
1s ↔ u(t) Y(s) = 1
s ·X(s) ↔ y(t) =∫ t
0 x(τ )dτ
1s+a ↔ e−at Y(s) = 1
s+a ·X(s) ↔ y(t) + ay(t) = x(t)
ωs2+ω2 ↔ sin(ωt) Y(s) = ω
s2+ω2 ·X(s) ↔ y(t) + ω2y(t) = ωx(t)
e−sT ↔ δ(t− T ) Y(s) = e−sT ·X(s) ↔ y(t) = x(t− T )
Transfer functions arise through Laplace Analysis of
Differential Equations which represent the dynamic behaviour
of systems. They summarise the input/output behaviour of a
system in the s domain (i.e. they generalize the concept of ‘gain’).
Finding the Laplace transform of a signal is a step one takes when
it is required to find out what a particular system (represented by its
transfer function) does to an input signal.
It just turns out that the Laplace transform of some signals have a
form which occurs also as a Transfer function of particular systems.
3C1 Signals and Systems 20 www.mee.tcd.ie/∼sigmedia
8 MORE BLOCK DIAGRAM ALGEBRA
Terminology (will be used later in examining stability)
Suppose
G(s) =n(s)
d(s)(11)
Then the roots of n(s) are called the ZEROS of the system G(s)
And the roots of d(s) are called the POLES of the system G(s)
EXAMPLE:
G(s) =4s2 − 8s− 60
s3 + 2s2 + 2s
=4(s + 3)(s− 5)
s(s + 1 + j)(s + 1− j)
Zeros of G(s) are the values of s that make the numerator = 0 hence
Poles of G(s) are the values of s that make the denominator = 0
hence
3C1 Signals and Systems 21 www.mee.tcd.ie/∼sigmedia
8.1 DC Motor 8 MORE BLOCK DIAGRAM ALGEBRA
8.1 Another Example: FIELD CONTROLLED DC MOTOR
Left: Circuit Diagram, Right: The Motor
DC Motors found in Hard Drives, CD Players, Scalextric Model car sets
How does θ(t) (the rotational angle of the axle) depend on e(t) (the voltage
applied to the field winding)? The following system equations apply:
Linearised Motor Equation (k=some const.)
τ(t) = ki(t) (1)
Torque = Inertia × acceleration
τ(t)−Bθ(t) = Jθ(t) (2)
Kirchoff
e(t) = Ri(t) + L∂i
∂t(3)
1. What is the the system transfer function θ(s)E(s)?
2. What is the impulse response of the system?
3. What is the response of the system to a unit step impulse at t = 0?
Assume θ(0) = 0, θ(t) = 0, i(0) = 0; First have a think about what you
3C1 Signals and Systems 22 www.mee.tcd.ie/∼sigmedia
8.1 DC Motor 8 MORE BLOCK DIAGRAM ALGEBRA
expect . . .
Now some analysis:
• To get the transfer function we need to express the Laplace transform of
the output signal τ(s) (torque) as a function of the laplace transform of
the input signal I(s) (current). So take Laplace xforms of the
differential equations representing the dynamic behaviour of each
system (remember tables!) . . .
From eqn 1
⇒ τ (s) = KI(s) (4)
From eqn. 2
⇒ τ(s)−B[sΘ(s)− θ(0)] = J [s2Θ(s)− sθ(0)− (θ)(0)]
But from initial conditions, θ(0) = 0, θ(t) = 0, i(0) = 0 so
τ(s) = Θ(s)[Bs + Js2] (5)
From eqn. 3
E(s) = RI(s) + L[sI(s)− i(0)] (6)
= I(s)[R + Ls]
3C1 Signals and Systems 23 www.mee.tcd.ie/∼sigmedia
8.1 DC Motor 8 MORE BLOCK DIAGRAM ALGEBRA
Let Tf = L/R, Tm = J/B (to reduce the amount of writing to do).
Hence from eqn. 6
I(s) =E(s)
R(1 + sTf)(7)
Still have some work to do ... don’t panic ... all we want to do is to get
Θ(s) = somefunctionof(E(s))
Subst I from eqn 7 into 4
⇒ τ(s) =
(K
R(1 + sTf)
)E(s) (8)
Subst τ from 8 into 5
K
R(1 + sTf)E(s) = Θ(s)[Bs + BTms2]
= Θ(s)Bs[1 + Tms]
So
Θ(s)
E(s)=
K
R(1 + sTf)
1
Bs[1 + Tms]
Required Transfer Function is
Θ(s)
E(s)=
K/(RB)
s(1 + sTf)(1 + Tms)(9)
We shall denote this system transfer function as H(s).
• Impulse Response: Put an impulse as input into the system and
calculate the output. Therefore set e(t) = δ(t) ⇒ E(s) = 1 Hence, to
calculate output (which is the impulse response when an impulse is
input)
Θ(s) = H(s)E(s) = H(s)× 1
= H(s) (10)
3C1 Signals and Systems 24 www.mee.tcd.ie/∼sigmedia
8.1 DC Motor 8 MORE BLOCK DIAGRAM ALGEBRA
Therefore, the time domain impulse response h(t) is just L−1(H(s))
Remember: System transfer function is Laplace transform of the
impulse response, or the impulse response is the inverse Laplace
Transform of the system transfer function. So
h(t) = L−1
K/(RB)
s(1 + sTf)(1 + Tms)
(11)
Need to use partial fractions to express the transfer function in a
simpler form to make taking the inverse easier (i.e. to use tables).
K/(RB)
s(1 + sTf)(1 + Tms)=
A
s+
B
(1 + sTf)+
C
(1 + sTm)
Using “Cover Up” rule
A =K/(RB)
(1 + 0× Tf)(1 + Tm × 0)=
K
RB
B =K/(RB)
s(1 + Tms)
∣∣∣∣s=− 1
Tf
=K/(RB)
− 1Tf
(1− TmTf
)
=−KTf/(RB)
(1− TmTf
)
=−KT 2
f /(RB)
(Tf − Tm)
C =K/(RB)
s(1 + sTf)
∣∣∣∣s=− 1
Tm
=K/(RB)
− 1Tm
(1− Tf
Tm)
=−KTm/(RB)
(1− Tf
Tm
=KT 2
m/(RB)
(Tf − Tm)
Hence (after some simplification)
K/(RB)
s(1 + sTf)(1 + Tms)=
K
RB
[1
s− Tf
(Tf − Tm)
(1
s + 1Tf
)+
Tm
Tf − Tm
(1
s + 1Tm
)]
3C1 Signals and Systems 25 www.mee.tcd.ie/∼sigmedia
8.1 DC Motor 8 MORE BLOCK DIAGRAM ALGEBRA
So now we can use tables to find the inverse Laplace Transform, hence
H(s) =K
RB
[1
s− Tf
(Tf − Tm)
(1
s + 1Tf
)+
Tm
Tf − Tm
(1
s + 1Tm
)]
⇒ h(t) = L−1 H(s)
=K
RB
[u(t)− Tf
(Tf − Tm)e−
(1
Tft
)
+Tm
Tf − Tme−( 1
Tmt)
]
Consider:
– Tf generally smaller than Tm.
– Electronic cct transient FAST, Mechanical cct transient SLOW
• Step response g(t) (say) : can get this either i) integrate the impulse
response OR ii) set e(t) = u(t) and then calculate output. You can spot
that the only terms that involve t in h(t) have simple exponentials, so
integration seems straightforward
g(t) =
∫ t
0h(τ)dτ
=K
RB
[∫ t
0
(u(τ)− Tf
(Tf − Tm)e−
(1
Tfτ)
+Tm
Tf − Tme−( 1
Tmτ)
)dτ
]
remember∫
e−t/a = 1−(1/a)e
−t/a = −ae−t/a
=K
RB
[τ +
T 2f
(Tf − Tm)e−
(τ
Tf−1
)
− T 2m
Tf − Tme−( τ
Tm−1)
]t
0
=K
RB
[t +
T 2f
(Tf − Tm)e−
(t
Tf−1
)− T 2
m
Tf − Tme−( t
Tm−1)
]
Steady state response is g(t) as t →∞. So steady state response is
gt→∞(t) =Kt
RB(12)
3C1 Signals and Systems 26 www.mee.tcd.ie/∼sigmedia