Chapter 11
Laplace transforms and CT LTI systems
We have seen that the Laplace transform allows us to have a frequency domain-like represen- tation of signals, even if they are not absolutely summable. By taking the Laplace transform of the impulse response of an LTI system, we obtain a generalization of the frequency response of a CT LTI system that is valid even if the system is not BIBO stable. This representation is called the transfer function of an LTI system. This is particularly important in the study of feedback control systems, where many of the systems of interest are unstable.
In this topic we introduce the transfer function, and show how to use the Laplace transform and the inverse Laplace transform to find the response of a CT LTI system to a given input. We conclude the chapter with a discussion of the one-sided Laplace transform. This is a slight modification of the Laplace transform that takes into account the initial conditions of a system. It allows us to extend our Laplace-domain analysis techniques to systems described by linear, constant coefficient, differential equations that do not satisfy the initial rest condition.
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11.1 Transfer functions of LTI systems
Just as the frequency response of an LTI system is related to the system output when the input is a unit complex exponential x(t) = ejω0t, the transfer function of an LTI system is related to the system output when the input is a general complex exponential x(t) = est.
Response of LTI system to general complex exponential
Let s be a complex number and let x(t) = est be a general complex exponential signal. If an LTI system has impulse response h and input x then there is a function Hˆ such that
the output is
We call the complex-valued function Hˆ the transfer function of the system.
y(t) = (h ∗ x)(t) = Hˆ (s)est.
To see why this is true we use the convolution formula:
=e h(τ)e dτ −∞
= Hˆ(s)est.
Notice that Hˆ is precisely the Laplace transform of the impulse response.
The frequency response is the Fourier transform of the impulse response of a BIBO stable LTI system. The transfer function is the Laplace transform of the impulse response of an LTI system. We can use the relationship between the Fourier transform and the Laplace transform to relate the frequency response and the transfer function.
Note that this relation may remain true for some systems that are not BIBO stable (such as the integrator system) for which we defined a frequency response, even though the system was not BIBO stable.
One way to see why this holds is to use the convolution property of Laplace transforms. We know that the output y of the system is the convolution of the input x and the impulse response h, i.e., y(t) = (x ∗ h)(t) for all t. Taking the Laplace transform of both sides, and using the fact that the transfer function is the Laplace transform of the impulse response, we see that Yˆ (s) = Hˆ (s)Xˆ (s) for all s in both RoC(h) and RoC(x).
h(τ)x(t − τ) dτ s(t−τ)
h(τ)e dτ st ∞ −sτ
Transfer function and impulse response
The transfer function of an LTI system is the Laplace transform of the impulse response of the system.
Transfer function and frequency response
If Hˆ is the transfer function of a BIBO stable LTI system, and H is the frequency response
of the system then
H ( ω ) = Hˆ ( j ω ) .
Relating the input, output, and transfer function of LTI systems
uppose an LTI system has transfer function Hˆ . If Xˆ is the Laplace transform of the input of the system and Yˆ is the Laplace transform of the output of the system then
Yˆ(s)=Hˆ(s)Xˆ(s) foralls.
Example 11.1: Transfer function of delay
Forarealnumberτ,considerthesystemDelayτ definedbyDelayτ(x)(t)=x(t−τ). We can compute the transfer function by finding the output when the input is x(t) = est. This is
x(t − τ) = es(t−τ) = e−τsest.
Hence the transfer function is
and the region of convergence is RoC(h) = C. Note that this is not a rational function.
Hˆ ( s ) = e − τ s
Example 11.2: Transfer function of integrator system
Recall that the integrator system has output y defined by t
when the input is x. Recall that the impulse response of the system is
the unit step function. To find the transfer function, we will take the Laplace transform of the unit step function. From topic 10 we know that the Laplace transform of e−αtu(t), for any complex number α, is 1/(s + α). Putting α = 0 we see that the Laplace transform of the unit step, and hence the transfer function of the integrator system, is
Hˆ ( s ) = 1s
and the region of convergence is {s ∈ C : Re(s) > 0}.
δ(τ) dτ = u(t),
Example 11.3: Transfer function of differentiator system
The differentiator system has output y defined by y(t) = dx when the input is x. To find dt st
the transfer function, we can compute the output when the input is x(t) = e . This is y(t) = sest
and so the transfer function of the differentiator is Hˆ (s) = s.
This is a rational transfer function with a pole at infinity. So the region of convergence is the whole complex plane (except for the ‘point at infinity’). Worrying about what happens at infinity might seem pedantic, but it is important when we assess stability and causality via the region of convergence.
11.1.1 Transfer function of compositions of LTI systems
In Topic 6 we saw that when we interconnection LTI systems, we can find the frequency response of the overall system in a fairly straightforward way. Very similar results hold for the transfer functions of interconnections of LTI systems.
Series interconnection
If two LTI systems are connected in series, then the transfer function is the product of the transfer functions of the individual systems.
est Hˆ 1 (s)est Hˆ 1 (s)Hˆ 2 (s)est
Transfer function of series interconnection
If S1 and S2 are LTI systems with transfer functions Hˆ1 and Hˆ2 then the frequency response of the series interconnection of S1 and S2 is
Hˆ (s) = Hˆ1(s)Hˆ2(s).
To see why, recall that the impulse response h of the series composition of two LTI systems is the convolution of the impulse responses, h1 and h2, of the two systems. Using the convolution property of the Laplace transform, and the fact that the transfer function is the convolution of the impulse response, we see that Hˆ(s) = Hˆ1(s)Hˆ2(s).
Parallel interconnection
If two LTI systems are connected in parallel, then the transfer function is the sum of the transfer function of the individual systems.
Hˆ 1 (s)est
Hˆ 2 (s)est
(Hˆ1(s) + Hˆ2(s))est
Transfer function of parallel interconnection
If S1 and S2 are LTI systems with transfer functions H1 and H2 then the transfer function of the parallel interconnection of S1 and S2 is
Hˆ ( s ) = Hˆ 1 ( s ) + Hˆ 2 ( s ) .
To see why, simply note that if x(t) = est is the input to the system, the output of the first system is y1(t) = Hˆ1(s)est and the output of the second system is y2(t) = Hˆ2(s)est. The overall output of the system is
y(t) = y1(t) + y2(t) = (Hˆ1(s) + Hˆ2(s))est = Hˆ (s)est. 4
Feedback interconnection
We now find the transfer function of the feedback interconnection of two LTI systems S1 and S2 shown below
Transfer function of feedback interconnection
Suppose S1 and S2 are LTI systems with transfer functions Hˆ1 and Hˆ2. If S1 and S2 are connected in feedback as shown above. Then the transfer function of the overall system is
Hˆ ( s ) = Hˆ 1 ( s ) . 1 − Hˆ1(s)Hˆ2(s)
To see why this is true, let Yˆ denote the Laplace transform of y and let Xˆ denote the Laplace transform of x. Now we can see that
Eliminating Zˆ gives Rearranging this we see that
Yˆ(s) = Hˆ1(s)(Xˆ(s) + Zˆ(s)) Zˆ(s) = Hˆ2(s)Yˆ (s).
Yˆ(s) = Hˆ1(s)(Xˆ(s) + Hˆ2(s)Yˆ(s)). Yˆ(s)(1 − Hˆ1(s)Hˆ2(s)) = Hˆ1(s)Xˆ(s).
Finally we use the fact that the transfer function of the overall system is Hˆ ( s ) = Yˆ ( s )
Hˆ ( s ) = Hˆ 1 ( s ) .
1 − Hˆ1(s)Hˆ2(s)
Summary of Section 11.1
If the input to a continuous-time LTI system is est (where s is a complex number), the output is Hˆ(s)est. The function Hˆ is called the transfer function of the system. The transfer function is the Laplace transform of the impulse response of the system. For BIBO stable systems, the transfer function and the frequency response are related by
H(ω) = Hˆ (jω). If an LTI system has transfer function Hˆ and input with Laplace transform Xˆ then the output has Laplace transform Yˆ (s) = Hˆ (s)Xˆ (s). Finally, when we interconnect subsystems in series, parallel, or feedback, the transfer function of the overall system is related (in a fairly simple way) to the transfer function of the subsystems.
11.2 The region of convergence and properties of LTI systems
We have seen that there are close connections between the structure of the region of conver- gence of a Laplace transform and whether the associated signal is left- or right- or two-sided, and whether it is absolutely integrable. For LTI systems, there are close connections between whether a system is causal or BIBO stable, and the region of convergence of the Laplace trans- form.
Region of convergence and causal systems
If an LTI system is causal then its impulse response is right-sided. It follows that
• for a causal LTI system the region of convergence of the transfer function has the form
{s ∈ C : Re(s) > a}
for some real number a.
• for a causal LTI system with a rational transfer function the region of convergence is
{s ∈ C : Re(s) > Re(p)} where p is the rightmost pole of the transfer function.
Region of convergence and BIBO stability
An LTI system is BIBO stable if and only if its impulse response if absolutely integrable. It follows that an LTI system is BIBO stable if and only if the region of convergence of its transfer function contains the imaginary axis.
Region of convergence for causal BIBO stable systems
• A causal LTI system is BIBO stable if and only if the region of convergence of its transfer function is of the form
{s ∈ C : Re(s) > a} where a is a negative real number.
• A causal LTI system with rational transfer function is BIBO stable if and only if all of its poles are have negative real part.
Example 11.4: Stability of integrator system
Recall from Example 11.2 that the transfer function of the integrator system is Hˆ (s) = 1/s with region of convergence {s ∈ C : Re(s) > 0}. The integrator is not a BIBO stable
system. One way to see this is that the region of convergence of the transfer function does not contain the imaginary axis.
The following example illustrates the subtlety that arises when we have a pole ‘at infinity’. This notion and the mathematics associated with it can be made rigorous, but it is well beyond the scope of this unit to do so.
Example 11.5: Stability of differentiator system
Recall from Example 11.3 that the transfer function of the differentiator system is Hˆ (s) = s with region of convergence being the complex plane except for the ‘point at infinity’ where it has a pole.
The differentiator system is not BIBO stable. One way to see this is to consider the input x(t) = sin(et). This is clearly a bounded signal, because |x(t)| ≤ 1 for all t. Its
derivative is
dx = et cos(et) dt
which is unbounded.
The pole at infinity means that the region of convergence does not contain the imagi-
nary axis (since the imaginary axis contains the ‘point at infinity’). This is consistent with the system not being BIBO stable.
Summary of Section 11.2
The region of convergence of the transfer function is closely related to properties of the corresponding LTI system. For instance, the region of convergence contains the imaginary axis if and only if the system is BIBO stable, and the region of convergence is to the right of the right-most pole if and only if the system is causal.
11.3 Inverse Laplace transforms and computing system response
We have two options to compute the system response for general (not necessarily absolutely summable) input signals and general (not necessarily BIBO stable) LTI systems. The first is based on convolution:
The second is based on exploiting the convolution property of the Laplace-transform
Time-domain approach
1. Find the impulse response.
2. Compute the convolution of the impulse response and the input.
Laplace transform-based approach
1. Find the transfer function Hˆ.
2. Find the Laplace transform Xˆ of the input.
3. ComputetheLaplacetransformYˆ(s)=Hˆ(s)Xˆ(s)oftheoutput.
4. Find the signal y with Laplace transform Yˆ (s).
In this section we focus on the last step of the Laplace transform-based approach. Given a Laplace transform and its region of convergence, how can we find the signal with that Laplace transform? This process is called finding the inverse Laplace transform.
There is a formula for the inverse Laplace transform, but it involves integration in the complex plane, and is not so useful in practice. In practice, we usually compute inverse Laplace transforms by decomposing a Laplace transform as a sum of simple pieces, ‘looking up’ the inverse Laplace transforms of the pieces in a table (possibly also using other Laplace transform properties), and then using linearity.
One of the challenges inherent in this approach is that the algebraic expression for the Laplace transform does not specify the inverse Laplace transform completely. We also need to consider the region of convergence.
Example 11.6
Xˆ(s)= 1 − 1 , s+1 s+2
with region of convergence {s ∈ C : Re(s) > −1}. Note that this signal has poles at s = −1 and s = −2. To find the inverse Laplace transform, we consider each of the terms in the sum separately.
• For 1 we know that the inverse Laplace transform could be either the right-sided
s+1−t −t signal e u(t) or the left-sided signal −e
• For 1 we know that the inverse Laplace transform could be either the right-sided
signal e−2tu(t)or the left-sided signal −e−2tu(−t).
To determine which (of the four possible) combinations could be the inverse Laplace trans- form of Xˆ we must consider the region of convergence. Since the region of convergence of Xˆ istotheright ofboththepoleats=−1andthepoleats=−2,wecandeducethat the correct inverse Laplace transform for each term is the right-sided one. In other words, the inverse Laplace transform of Xˆ is
x(t) = e−tu(t) − e−2tu(t).
Example 11.7
We consider the same example, Xˆ (s) = 1 − 1 but this time with region of convergence s+1 s+2
{s ∈ C : −2 < Re(s) < −1}. Thisistotheleftofthepoleof 1 ats=−1andtotherightofthepoleof 1 at
s = −2. That means that we should take the left-sided inverse Laplace transform of the
first term, and the right-sided inverse Laplace transform of the second term. Overall, we find that the inverse Laplace transform is
x(t) = −e−tu(−t) − e−2tu(t).
In the following example, we use properties of the Laplace transform to help us identify the inverse Laplace transform of each term.
Example 11.8
Consider the transfer function
Hˆ(s)= es s+1
with region of convergence {s ∈ C : Re(s) > −1}. The inverse Laplace transform will tell us the impulse response of the system. To compute this, we first recognise that Hˆ(s) = esXˆ(s)
Xˆ(s)= 1 . s+1
The inverse Laplace transform of Xˆ (s) (with region of convergence Re(s) > −1) is x(t) = e−tu(t).
Recall that the transfer function of the system Delay−1 is precisely es. This means that h(t) is the output when we pass x(t) through the system Delay−1. In other words,
h(t) = Delay−1(x)(t) = x(t + 1) = e−(t+1)u(t + 1).
If Xˆ(s) is a rational function, we can decompose it as a sum of ‘simple’ rational functions using partial fractions. We can then look up the inverse Laplace transform of each of the pieces in a table. This gives a systematic procedure to compute the inverse Laplace transform of any rational function. In this section, we describe this procedure in the case where all the poles have multiplicity one, and merely sketch what happens in the general case.
11.3.1 Strictly proper and all poles of multiplicity one
Assume that Xˆ(s) is rational, strictly proper (i.e., the degree of the numerator is strictly smaller than the degree of the denominator) and all the poles have multiplicity one. Then we can rewrite Xˆ(s) in the following way:
1. Factorise the denominator to find the poles:
Xˆ ( s ) = B ( s ) = B ( s )
A(s) (s−p1)···(s−pn)
so that the (possibly complex) poles are at p1, p2, . . . , pn and are all distinct.
2. Find numbers (possibly complex) A1, A2, . . . , An such that Xˆ(s)= B(s) =nAi.
(s−p1)···(s−pn) i=1 s−pi
This is called a partial fraction decomposition of Xˆ(s). We can find the coefficients by using
the following formula:
Ai=lim(s−pi)Xˆ(s) fori=1,2,…,n. s→pi
Once we have the Laplace transform in partial fraction form, we can find the inverse Laplace transform term-by-term, being sure to choose either the left- or right-sided inverse transform for each term based on whether the region of convergence is to the left or right of the pole corresponding to that term.
Example 11.9
Consider an LTI system with strictly proper rational transfer function Hˆ ( s ) = s − 1
(s+1)(s−2)
and region of convergence {s ∈ C : Re(s) > 2}. The inverse Laplace transform is the impulse response of the system. To find this, we first decompose Hˆ (s) into partial fractions
To find A1, which is associated with the pole at s = −1, we evaluate
s−1 =A1+A2. (s+1)(s−2) s+1 s−2
s−1 2 A1=lim(s+1)Hˆ(s)= =.
s→−1 s−2s=−1 3
Notice how we could evaluate the limit by substitution since the term (s + 1) cancels with
the pole in Hˆ(s). Similarly
A2 =lim(s−2)Hˆ(s)= = .
s−1 1 s→2 s+1s=2 3
Hˆ ( s ) = 2 1 + 1 1 . 3s+1 3s−2
Since the region of convergence is to the right of both of the poles, we should take the right-sided inverse Laplace transform of both of the terms. This gives an impulse response
h(t) = 23e−tu(t) + 13e2tu(t).
11.3.2 Not strictly proper
If the given Laplace transform Xˆ(s) is a rational function that is not strictly proper (i.e., the numerator has degree that is greater than or equal to the denominator), we first need to perform polynomial division. (This is the generalisation to polynomials of rewriting an improper fraction such as 32 as the sum of an integer plus a proper fraction 32 = 1 + 12 .)
Example 11.10
Consider the Laplace transform
Xˆ ( s ) = s 2 + 3 s + 1 s+1
with region of convergence {s ∈ C : Re(s) > −1}. The degree of the numerator is not strictly smaller than the degree of the denominator, so we first perform division. By making a common denominator, it is straightforward to check that
Xˆ(s)=s2+3s+1=s+2− 1 . s+1 s+1
We can now compute the inverse Laplace transform. The term corresponding to the pole
at s = −1 will have right-sided inverse Laplace transform e−tu(t) because the region of
convergence of Xˆ is to the right of the pole at s = −1. The inverse Laplace transforms of
the constant 2 is 2δ(t). The inverse Laplace transform of s is d δ(t), which can be found dt
by using the differentiation property of the Laplace transform. Putting this together we
x(t)= dδ(t)+2δ(t)−e−tu(t). dt
11.3.3 Poles of higher multiplicity
If the given Laplace transform has poles that do not all have multiplicity one, finding the partial fraction decomposition is more complicated. In general, suppose Xˆ is strictly proper and
Xˆ ( s ) = B ( s ) (s−p1)m1(s−p2)m2 ···(s−pk)mk
where mi is the multiplicity of pole pi. Then Xˆ has a partial fraction decomposition of the form
k Ri1 Ri2 Xˆ ( s ) = +
Rim + · · · + k
. NowtofindthisdecompositionweneedtofindthecoefficientsRij for1≤i≤kand1≤j≤mi.
i=1 (s−pi) (s−pi)2 We will not discuss how to do this in this unit.
Example 11.11
Consider the Laplace transform
Xˆ(s)= s3 +s2 +s+1. s(s + 2)2
Suppose it has region of convergence
{s ∈ C : −2 < Re(s) < 0}.
This has a pole of multiplicity two at s = −2 and it is not strictly proper (since the numer- ator has degree three, which is not strictly smaller than the degree of the denominator).
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