Home/Chapter 6

Frequency Domain Analysis of Linear Time-Invariant (LTI) Systems

Input-output relations for LTI systems via Fourier analysis. Computing transfer functions for convolution systems using Fourier transforms.

Input-Output Relations for Linear Time-Invariant Systems via Fourier Analysis

As we discussed in the relevant section, a very important property of convolution systems is that if the input is a harmonic function, so is the output.

Continuous-Time Case. Let uL(R;C)u \in L_\infty(\mathbb{R}; \mathbb{C}) given with

u(t)=ei2πft,u(t) = e^{i2\pi f t},

be the input to a linear time-invariant system

y(t)=τ=h(tτ)u(τ)dτ=τ=h(τ)u(tτ)dτy(t) = \int_{\tau=-\infty}^{\infty} h(t - \tau) u(\tau)\, d\tau = \int_{\tau=-\infty}^{\infty} h(\tau) u(t - \tau)\, d\tau

Then, the output satisfies

y(t)=(h(s)ei2πfsds)ei2πfty(t) = \left(\int_{-\infty}^{\infty} h(s) e^{-i2\pi f s}\, ds\right) e^{i2\pi f t}

The integral

h^(f):=(h(t)ei2πftdt),\hat{h}(f) := \left(\int h(t) e^{-i2\pi f t}\, dt\right),

is FCC(h)\mathcal{F}_{CC}(h) evaluated at ff. We call this value, the frequency response of the system at frequency ff, whenever it exists.

Remark.

The frequency response h^(f)\hat{h}(f) tells us how the system scales and phase-shifts a pure sinusoid at frequency ff. The magnitude h^(f)|\hat{h}(f)| gives the gain and h^(f)\angle \hat{h}(f) gives the phase shift. This is the fundamental connection between the time-domain description (impulse response hh) and the frequency-domain description (frequency response h^\hat{h}) of an LTI system.

Discrete-Time Case. A similar discussion applies for a discrete-time system: Let hl1(Z;C)h \in l_1(\mathbb{Z}; \mathbb{C}). If u(n)=ei2πfnu(n) = e^{i2\pi f n} is the input to a linear time-invariant system given with

y(n)=m=h(nm)u(m)=m=h(m)u(nm).y(n) = \sum_{m=-\infty}^{\infty} h(n - m) u(m) = \sum_{m=-\infty}^{\infty} h(m) u(n - m).

then

y(n)=(m=h(m)ei2πfm)ei2πfn.y(n) = \left(\sum_{m=-\infty}^{\infty} h(m) e^{-i2\pi f m}\right) e^{i2\pi f n}.

We recognize that

h^(f):=m=h(m)ei2πfm=(FDC(h))(f),\hat{h}(f) := \sum_{m=-\infty}^{\infty} h(m) e^{-i2\pi f m} = \left(\mathcal{F}_{DC}(h)\right)(f),

and call h^\hat{h} the frequency response function.

Convolution systems are used as filters through the characteristics of the frequency response.

Some Properties.

Recall the following properties of FCC\mathcal{F}_{CC}.

(i) Convolution Property. Let u,vSu, v \in \mathcal{S}. We have that

(FCC(uv))(f)=u^(f)v^(f)\left(\mathcal{F}_{CC}(u * v)\right)(f) = \hat{u}(f)\hat{v}(f)
Remark.

This is the most important property for LTI system analysis: convolution in time becomes multiplication in frequency. If y=huy = h * u, then y^(f)=h^(f)u^(f)\hat{y}(f) = \hat{h}(f)\hat{u}(f). This means the output spectrum is simply the input spectrum scaled by the frequency response at each frequency -- a tremendous simplification compared to evaluating the convolution integral directly.

(ii) Differentiation Property. If v=ddtuv = \frac{d}{dt}u, then v^(f)=i2πfu^(f)\hat{v}(f) = i2\pi f\, \hat{u}(f).

(iii) Time-Shift Property for FDC\mathcal{F}_{DC}. Let v=σθ(u)v = \sigma^\theta(u) for some θZ\theta \in \mathbb{Z}, that is v(n)=u(n+θ)v(n) = u(n + \theta). Then,

v^(f)=nZu(n+θ)ei2πfn=ei2πθfu^(f)\hat{v}(f) = \sum_{n \in \mathbb{Z}} u(n + \theta) e^{-i2\pi f n} = e^{i2\pi \theta f}\, \hat{u}(f)

The above will be very useful properties for studying LTI systems. We can also obtain converse differentiation properties, which will be considered in further detail in the relevant section while studying the Z and the Laplace transformations. Nonetheless, we will present two such properties in the following (see Section A.2 for a justification on changing the order of differentiations and summations/integrations):

(iv) Differentiation in Frequency for FDC\mathcal{F}_{DC}. Let

FDC(x)(f)=x^(f)=nx(n)ei2πfn\mathcal{F}_{DC}(x)(f) = \hat{x}(f) = \sum_n x(n) e^{-i2\pi f n}

Then, through changing the order of limit and summation:

dx^(f)df=ndx(n)ei2πfndf=n(2iπnx(n))ei2πfn\frac{d\hat{x}(f)}{df} = \sum_n \frac{dx(n) e^{-i2\pi f n}}{df} = \sum_n (-2i\pi n\, x(n)) e^{-i2\pi f n}

This leads to the conclusion that with v(n)=nx(n)v(n) = -n\, x(n), with vv absolutely summable,

FDC(v)(f)=1i2πdx^(f)df\mathcal{F}_{DC}(v)(f) = \frac{1}{i2\pi} \frac{d\hat{x}(f)}{df}

(v) Differentiation in Frequency for FCC\mathcal{F}_{CC}. Likewise, for the continuous-time case with xSx \in \mathcal{S}

FCC(x)(f)=x^(f)=tx(t)ei2πftdt\mathcal{F}_{CC}(x)(f) = \hat{x}(f) = \int_t x(t) e^{-i2\pi f t}\, dt

Via the analysis in Section A.2, through changing the order of limit and integration,

dx^(f)df=dx(t)ei2πftdfdt=(2iπtx(t))ei2πtdt\frac{d\hat{x}(f)}{df} = \int \frac{dx(t) e^{-i2\pi f t}}{df}\, dt = \int (-2i\pi t\, x(t)) e^{-i2\pi t}\, dt

This leads to the conclusion that with v(t)=tx(t)v(t) = -t\, x(t), with v(t)v(t) (absolutely) integrable,

FCC(v)(f)=1i2πdx^(f)df\mathcal{F}_{CC}(v)(f) = \frac{1}{i2\pi} \frac{d\hat{x}(f)}{df}

Useful Transform Pairs. In the context of LTI systems, we will occasionally build on the following properties:

  • If u(t)=1{t0}eatu(t) = 1_{\{t \geq 0\}} e^{at}, with a<0a < 0, then u^(f)=1a+i2πf\hat{u}(f) = \frac{1}{-a + i2\pi f}.
  • Likewise for FDC\mathcal{F}_{DC}, for a<1|a| < 1, if u(n)=an11{n1}u(n) = a^{n-1} 1_{\{n \geq 1\}}, then u^(f)=ei2πf11aei2πf\hat{u}(f) = e^{-i2\pi f} \frac{1}{1 - a e^{-i2\pi f}}.

The properties above are crucial, and typically sufficient, for studying a large class of linear time invariant systems described by differential and difference equations (convolution systems).

Transfer Functions and their Computation for Convolution Systems via Fourier Transforms

In applications for control, communications, and signal processing, one may design systems or filters using the properties of the frequency response functions.

Continuous-Time Systems

Consider the following continuous-time system with input uu and output yy:

k=0Nakdkdtky(t)=m=0Mbmdmdtmu(t)\sum_{k=0}^{N} a_k \frac{d^k}{dt^k} y(t) = \sum_{m=0}^{M} b_m \frac{d^m}{dt^m} u(t)

Taking the FCC\mathcal{F}_{CC} of both sides, we obtain

(k=0Nak(i2πf)k)y^(f)=(m=0Mbm(i2πf)m)u^(f)\left(\sum_{k=0}^{N} a_k (i2\pi f)^k\right) \hat{y}(f) = \left(\sum_{m=0}^{M} b_m (i2\pi f)^m\right) \hat{u}(f)

This leads to:

h^(f)=m=0Mbm(i2πf)mk=0Nak(i2πf)k\hat{h}(f) = \frac{\sum_{m=0}^{M} b_m (i2\pi f)^m}{\sum_{k=0}^{N} a_k (i2\pi f)^k}
Remark.

The transfer function h^(f)\hat{h}(f) is a rational function of frequency. The numerator polynomial comes from the input side of the differential equation, and the denominator from the output side. This algebraic relationship is far easier to work with than the original differential -- system analysis reduces to algebraic manipulation of polynomials.

ExampleFirst-Order Continuous-Time System

As an example, let us consider

dydt=ay(t)+u(t),a>0\frac{dy}{dt} = -ay(t) + u(t), \quad a > 0

For this system, we obtain by taking the FCC\mathcal{F}_{CC} of both sides (assuming this exists), we have

h^(f)=1a+i2πf\hat{h}(f) = \frac{1}{a + i2\pi f}

which is consistent with a direct computation, as done earlier, of the Fourier transform of

h(t)=eat1{t0}.h(t) = e^{-at} 1_{\{t \geq 0\}}.

Discrete-Time Systems

Likewise, for discrete-time systems:

k=0Naky(nk)=m=0Mbmu(nm)\sum_{k=0}^{N} a_k\, y(n-k) = \sum_{m=0}^{M} b_m\, u(n-m)

Taking the FDC\mathcal{F}_{DC} of both sides (assuming the FDC\mathcal{F}_{DC} exist), we obtain

h^(f)=y^(f)u^(f)=m=0Mbmei2πmfk=0Nakei2πkf\hat{h}(f) = \frac{\hat{y}(f)}{\hat{u}(f)} = \frac{\sum_{m=0}^{M} b_m e^{-i2\pi m f}}{\sum_{k=0}^{N} a_k e^{-i2\pi k f}}
ExampleFirst-Order Discrete-Time System

As an example, consider

y(n+1)=ay(n)+u(n),a<1y(n+1) = ay(n) + u(n), \quad |a| < 1

For this system, we obtain by taking the FDC\mathcal{F}_{DC} of both sides (assuming this exists), we arrive at (ei2πfa)y^(f)=u^(f)(e^{i2\pi f} - a)\hat{y}(f) = \hat{u}(f) and thus

h^(f)=1ei2πfa=ei2πf1aei2πf\hat{h}(f) = \frac{1}{e^{i2\pi f} - a} = \frac{e^{-i2\pi f}}{1 - a e^{-i2\pi f}}

Once again, as in the continuous-time setup, as discussed earlier, this is consistent with taking the FDC\mathcal{F}_{DC} of the impulse response function given by

h(n)=an11{n10}h(n) = a^{n-1} 1_{\{n-1 \geq 0\}}

Computing the Inverse Transform via Partial Fractions

How to compute the inverse transform? One can, using h^\hat{h}, compute h(t)h(t) or h(n)h(n), if one is able to compute the inverse transform. A useful method is the partial fraction expansion method. More general techniques will be discussed in the following chapter. Let

R(λ)=P(λ)Q(λ)=p0+p1λ++pMλMq0+q1λ++qNλN,λCR(\lambda) = \frac{P(\lambda)}{Q(\lambda)} = \frac{p_0 + p_1 \lambda + \cdots + p_M \lambda^M}{q_0 + q_1 \lambda + \cdots + q_N \lambda^N}, \quad \lambda \in \mathbb{C}

If M<NM < N, we call this fraction strictly proper. If MNM \leq N, the fraction is called proper and if M>NM > N, it is called improper.

If M>NM > N, we can write

R(λ)=T(λ)+R~(λ),R(\lambda) = T(\lambda) + \tilde{R}(\lambda),

where TT has degree MNM - N and R~(λ)\tilde{R}(\lambda) is strictly proper. We can in particular write:

R~(λ)=i=1K(k=1miAik(λλi)k)\tilde{R}(\lambda) = \sum_{i=1}^{K} \left(\sum_{k=1}^{m_i} \frac{A_{ik}}{(\lambda - \lambda_i)^k}\right)

where λi\lambda_i are the roots of QQ and mim_i is the multiplicity of λi\lambda_i.

This is important because we can use the expansion and the properties of FCC\mathcal{F}_{CC} and FDC\mathcal{F}_{DC} presented earlier in the chapter to compute the inverse transforms. Such approaches will be studied in further detail in the following chapter. In the following, we present two examples.

ExampleContinuous-Time System with Exponential Input

Consider a linear time invariant (LTI) system characterized by:

y(1)(t)=ay(t)+u(t),tRy^{(1)}(t) = -ay(t) + u(t), \quad t \in \mathbb{R}

with a>0a > 0.

a) Find the impulse response of this system.

b) Find the frequency response of the system.

c) Let u(t)=et1{t0}u(t) = e^{-t} 1_{\{t \geq 0\}}. Find y(t)y(t).

Solution. a-b) Taking the Fourier transform of the terms in the differential equation, we obtain

(i2πf+a)y^(f)=u^(f)(i2\pi f + a)\hat{y}(f) = \hat{u}(f)

Hence, the frequency response is:

h^(f)=1a+i2πf\hat{h}(f) = \frac{1}{a + i2\pi f}

The impulse response corresponds to the inverse Fourier transform of this, which is

h(t)=eat1{t0}h(t) = e^{-at} 1_{\{t \geq 0\}}

c) The function uu has its Fourier as:

u^(f)=11+i2πf\hat{u}(f) = \frac{1}{1 + i2\pi f}

Hence, the frequency response of the output will be

y^(f)=11+i2πf1a+i2πf=1a111+i2πf+11a1a+i2πf,\hat{y}(f) = \frac{1}{1 + i2\pi f} \cdot \frac{1}{a + i2\pi f} = \frac{1}{a - 1} \cdot \frac{1}{1 + i2\pi f} + \frac{1}{1 - a} \cdot \frac{1}{a + i2\pi f},

when a1a \neq 1. This leads to an inverse transform of:

y(t)=1a1et1{t0}+11aeat1{t0}y(t) = \frac{1}{a - 1} e^{-t} 1_{\{t \geq 0\}} + \frac{1}{1 - a} e^{-at} 1_{\{t \geq 0\}}

When a=1a = 1, we have

y^(f)=(11+i2πf)2\hat{y}(f) = \left(\frac{1}{1 + i2\pi f}\right)^2

Once we observe that (11+i2πf)2\left(\frac{1}{1+i2\pi f}\right)^2 is the derivative (1/(i2π))d11+i2πfdf-(1/(i2\pi)) \frac{d \frac{1}{1+i2\pi f}}{df}, and that differentiation in frequency domain leads to multiplication by i2πt-i2\pi t in time domain as was discussed in class, the result becomes (with a=1a = 1)

y(t)=teat1{t0}y(t) = t e^{-at} 1_{\{t \geq 0\}}
ExampleSecond-Order Discrete-Time System

Let a non-anticipative LTI system be given by:

y(n)=34y(n1)18y(n2)+u(n)y(n) = \frac{3}{4}y(n-1) - \frac{1}{8}y(n-2) + u(n)

a) Compute the frequency response of this system.

b) Compute the impulse response of the system.

c) Find the output when the input is u(n)=(12)n1{n0}u(n) = \left(\frac{1}{2}\right)^n 1_{\{n \geq 0\}}.

Solution. a) We obtain the frequency response with the following: We have

y^(f)(134ei2πf+18ei4πf)=u^(f)\hat{y}(f)\left(1 - \frac{3}{4}e^{-i2\pi f} + \frac{1}{8}e^{-i4\pi f}\right) = \hat{u}(f)

This leads to:

h^(f)=1134ei2πf+18ei4πf\hat{h}(f) = \frac{1}{1 - \frac{3}{4}e^{-i2\pi f} + \frac{1}{8}e^{-i4\pi f}}

b) We write the second degree polynomial (in terms of ei2πfe^{i2\pi f}) in the denominator, in terms of first degree polynomials as follows:

h^(f)=1134ei2πf+18ei4πf=1(112ei2πf)(114ei2πf)\hat{h}(f) = \frac{1}{1 - \frac{3}{4}e^{-i2\pi f} + \frac{1}{8}e^{-i4\pi f}} = \frac{1}{(1 - \frac{1}{2}e^{-i2\pi f})(1 - \frac{1}{4}e^{-i2\pi f})}

This can be written as

2112ei2πf+1114ei2πf\frac{2}{1 - \frac{1}{2}e^{-i2\pi f}} + \frac{-1}{1 - \frac{1}{4}e^{-i2\pi f}}

Hence, the output in time domain becomes:

h(n)=2(12)n1{n0}(14)n1{n0}h(n) = 2\left(\frac{1}{2}\right)^n 1_{\{n \geq 0\}} - \left(\frac{1}{4}\right)^n 1_{\{n \geq 0\}}

c) The input has its Fourier as

u^(f)=1112ei2πf\hat{u}(f) = \frac{1}{1 - \frac{1}{2}e^{-i2\pi f}}

Hence, by multiplication with h^(f)\hat{h}(f) above:

y^(f)=1(114ei2πf)(112ei2πf)2\hat{y}(f) = \frac{1}{(1 - \frac{1}{4}e^{-i2\pi f})(1 - \frac{1}{2}e^{-i2\pi f})^2}

This can be expanded as

y^(f)=A114ei2πf+B112ei2πf+C(112ei2πf)2\hat{y}(f) = \frac{A}{1 - \frac{1}{4}e^{-i2\pi f}} + \frac{B}{1 - \frac{1}{2}e^{-i2\pi f}} + \frac{C}{(1 - \frac{1}{2}e^{-i2\pi f})^2}

The values for A,B,CA, B, C can be shown to be equal to: A=1,B=2,C=2A = 1, B = -2, C = 2.

The first and the second terms above can be converted to the time-domain by the property that x(n)=rn1n0x(n) = r^n 1_{n \geq 0} gets mapped by FDC\mathcal{F}_{DC} to 11rei2πf\frac{1}{1 - r e^{-i2\pi f}} for r<1|r| < 1. For the last term, we observe that

C(112ei2πf)2=C2ddf(1(112ei2πf))ei2πfi2π\frac{C}{(1 - \frac{1}{2}e^{-i2\pi f})^2} = \frac{C}{2} \frac{d}{df}\left(\frac{1}{(1 - \frac{1}{2}e^{-i2\pi f})}\right) \frac{e^{i2\pi f}}{-i2\pi}

Using the fact that the derivative in the frequency domain leads to multiplication by (i2π)n(-i2\pi)n by the time domain, and that multiplication by ei2πfe^{i2\pi f} leads to a shift in the time-domain, we obtain that the contribution of the last term in time-domain would be

(n+1)(12)n+11{n+10}(n+1)\left(\frac{1}{2}\right)^{n+1} 1_{\{n+1 \geq 0\}}

Hence, the output becomes (by the observation that, derivative in frequency domain leads to a polynomial multiplication in the time domain)

y(n)=((14)n2(12)n)1{n0}+(n+1)(12)n+11{n+10}y(n) = \left(\left(\frac{1}{4}\right)^n - 2\left(\frac{1}{2}\right)^n\right) 1_{\{n \geq 0\}} + (n+1)\left(\frac{1}{2}\right)^{n+1} 1_{\{n+1 \geq 0\}}

Exercises

Problem 6.2.1RC Circuit Analysis

Consider the R-C circuit with the equations:

dVC(t)dt=1RCVC(t)+1RCu(t)\frac{dV_C(t)}{dt} = -\frac{1}{RC} V_C(t) + \frac{1}{RC} u(t)

a) Viewed as a linear time-invariant system, where uu is the input and VCV_C is the output, find the impulse response and the frequency response.

b) Qualitatively, plot the Bode diagram.

Problem 6.2.2RLC Circuit Analysis

Consider the R-L-C circuit with the dynamics

Ld2Qdt2+RdQdt+1CQ=u(t)L\frac{d^2Q}{dt^2} + R\frac{dQ}{dt} + \frac{1}{C}Q = u(t)

Note VC=Q/CV_C = Q/C.

a) Viewed as a linear time-invariant system, where uu is the input and VCV_C is the output, find the impulse response and the frequency response.

b) Qualitatively, plot the Bode diagram in the setup when RR is very small.

c) Show that when RR is very small, the ff value which maximizes the amplitude of the frequency response is around 12πLC\frac{1}{2\pi\sqrt{LC}}. Such a model is often used as an antenna of a radio receiver with the value of the capacitance denoting a tuning parameter.

Problem 6.3.1Continuous-Time System with Cosine Input

Consider a continuous-time system described by the equation:

dy(t)dt=ay(t)+u(t),tR,\frac{dy(t)}{dt} = ay(t) + u(t), \quad t \in \mathbb{R},

where a<0a < 0.

a) Find the impulse response of this system.

b) Suppose that the input to this system is given by cos(2πf0t)\cos(2\pi f_0 t). Let yf0y_{f_0} be the output of the system. Find yf0(t)y_{f_0}(t).

c) If exists, find

limf0yf0(t),\lim_{f_0 \to \infty} y_{f_0}(t),

for all tR+t \in \mathbb{R}_+.

Problem 6.3.2Discrete-Time System with Multiple Inputs

Let a system be described by:

y(n+1)=ay(n)+bu(n)+cu(n1),nZ.y(n+1) = ay(n) + bu(n) + cu(n-1), \quad n \in \mathbb{Z}.

a) For what values of a,b,ca, b, c is this system bounded-input-bounded-output (BIBO) stable?

b) Let a=2,b=1,c=1a = 2, b = 1, c = 1. Compute the impulse response of the system.

c) With a=2,b=1,c=1a = 2, b = 1, c = 1; find the output as a function of nn, when the input is u(n)=1{n0}u(n) = 1_{\{n \ge 0\}}.

Problem 6.3.3Continuous-Time System with Exponential Input

Consider a linear time invariant (LTI) system characterized by:

y(1)(t)=ay(t)+u(t),tRy^{(1)}(t) = -ay(t) + u(t), \quad t \in \mathbb{R}

with a>0a > 0.

a) Find the impulse response of this system.

b) Find the frequency response of the system.

c) Let u(t)=et1{t0}u(t) = e^{-t} 1_{\{t \ge 0\}}. Find y(t)y(t).

Problem 6.3.4Ideal Low-Pass Filter

Consider a continuous time LTI system with a frequency response

h^(f)=1{f<f0}fR\hat{h}(f) = 1_{\{|f| < f_0\}} \quad f \in \mathbb{R}

a) Find the impulse response of the system; that is the output of the system when the input is the signal representing the δ\delta distribution.

b) Find the CCFT of the output, when the input is given by

u(t)=etcos(f1t)1{t0}u(t) = e^{-t} \cos(f_1 t) 1_{\{t \ge 0\}}
Problem 6.3.5DCFT Computations

a) Let xl2(Z;C)x \in l_2(\mathbb{Z}; \mathbb{C}). Compute the DCFT of

x(n)=an1{n0},x(n) = a^n 1_{\{n \ge 0\}},

with a<1|a| < 1.

b) Compute the DCFT of xx:

x(n)=cos(3πf0n)x(n) = \cos(3\pi f_0 n)
Problem 6.3.6Multi-Dimensional CDFT

Many signals take values in multi-dimensional spaces. If you were to define a CDFT for signals in L2([0,P1]×[0,P2];C)L_2([0, P_1] \times [0, P_2]; \mathbb{C}), for given P1,P2R+P_1, P_2 \in \mathbb{R}_+, how would you define it?

Problem 6.3.7Second-Order Discrete-Time System

Let a non-anticipative LTI system be given by:

y(n)=34y(n1)18y(n2)+u(n)y(n) = \frac{3}{4}y(n-1) - \frac{1}{8}y(n-2) + u(n)

a) Compute the frequency response of this system.

b) Compute the impulse response of the system.

c) Find the output when the input is u(n)=(12)n1{n0}u(n) = \left(\frac{1}{2}\right)^n 1_{\{n \ge 0\}}.