MIT 6.007 Signals and Systems Course Notes 2

Visit original course website for more details.

Lecture 4: Convolution

How to decompose signals?

Decompose input signals into a linear combination of basic signals.

If we have a LTI System and the response of basic signals, it will be easy to compute the response of the input signals.

• Delayed Impulses – Convolution
• Complex Exponentials – Fourier Analysis

Discrete-time Case

Decomposition

\[x[n] = \sum_{k=-\infty}^{+\infty} x[k] \delta[n-k]\]

Linear System

\[y[n] = \sum_{k=-\infty}^{+\infty} x[k] h_k[n]\] \[\delta[n-k] \rightarrow h_k[n]\]

If Time-Invariant

\[h_k[n] = h[n-k]\]

• Convolution Sum

\[y[n] = \sum_{k=-\infty}^{+\infty} x[k] h[n-k] = x[n] * h[n]\]

Continuous-time Case

Decomposition

\[x(t) = \int_{-\infty}^{+\infty} x(\tau) \delta(t-\tau) \mathrm{d}\tau\]

LTI System

Similarly, we have the convolution integral here:

\[y(t) = \int_{-\infty}^{+\infty} x(\tau) h(t-\tau) \mathrm{d}\tau = x(t) * h(t)\]

Lecture 5: Properties of LTI Systems

In this lecture, the continuous-time case and the discrete-time case are similar. Thus we will use notations in a less strict way. Sometimes only continuous-time case will be discussed.

Commutative, Associative, Distributive Properties

• Commutative

\[x * h = h * x\]

We can interchange the roles of input and impulse response for LTI systems.

• Associative

\[x * (h_1 * h_2) = (x * h_1) * h_2\]

We can write \(x * h_1 * h_2\) unambiguously.

Associative property combined with the commutative property implies that the output of cascaded LTI systems is independent of the order in which the systems are cascaded.

• Distributive over addition

\[x * (h_1 + h_2) = x * h_1 + x * h_2\]

A parallel combination can be collapsed into a single system whose impulse response is the sum of the two individual ones.

A Useful Fact for Linear Systems (Zero in, Zero out)

For a linear system, if the input is zero for all \(t\), then the output is also zero for all \(t\).

Other properties

Memory

If an LTI system is memoryless, then the impulse response must be a scaled impulse.

Proof:

\(y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)\mathrm{d}\tau\) relies only on the value of \(x(\tau)\) at \(\tau = t\). Thus, \(h\) is non-zero only at \(\tau = t\), which indicates that \(h(t) = k\delta(t)\).

Invertibility

If an LTI system with impulse response \(h\) is invertible, the inverse system has the impulse response \(h^{-1}\) that satisfies \(h * h^{-1} = h^{-1} * h = \delta\).

Stability

For LTI systems an equivalent condition to stability is that the impulse response be absolutely summable (discrete time) or absolutely integrable (continuous time).

Causality

If an LTI system is causal, then its impulse response must be zero for t (or n) < 0; furthermore, if the impulse response has this property, then the system is guaranteed to be causal.

Proof:

For the first part, use the fact that zero input leads to zero output for linear systems. For the second part, a simple way to prove this is by doing integration(summation) from minus infinity to t (n) instead of from minus infinity to plus infinity.

Operation Definition of Impulse

Focus on how it behaves in convolution instead of what it looks like.

Impulse function can be defined as something that satisfies the following when doing convolution:

\[x * \delta = x\]

The k-th order derivative of the impulse function \(u_{k}(t)\) is defined as:

\[x(t) * u_{k}(t) = \frac{\mathrm{d}^k x(t)}{\mathrm{d}t^k}\]

The m-th order integral of the impulse function \(u_{-m}(t)\) is defined as:

\[x(t) * u_{-m}(t) = \text{m-th running integral}\]

Facts:

\[u_{0}(t) = \delta(t)\] \[u_{-1}(t) = u(t)\] \[u_{k}(t) * u_{l}(t) = u_{k+l}(t)\]

Lecture 6: Systems represented by differential and difference equations

Linear Constant-Coefficient Differential Equations

\[\sum_{k=0}^N a_k \frac{\mathrm{d}^k y(t)}{\mathrm{d}t^k} = \sum_{k=0}^M b_k \frac{\mathrm{d}^k x(t)}{\mathrm{d}t^k} \qquad\text{(1)}\]

Homogeneous Equation

\[\sum_{k=0}^N a_k \frac{\mathrm{d}^k y_h(t)}{\mathrm{d}t^k} = 0 \qquad\text{(2)}\]

Given \(x(t)\), if particular solution \(y_p(t)\) satisfies (1) and homogeneous solution \(y_h(t)\) satisfies (2), then \(y(t) = y_h(t) + y_p(t)\) is the solution to (1).

The form of homogeneous solution

Suppose \(y_h(t) = A\mathrm{e}^{st}\) is the solution, then we get:

\(\sum_{k=0}^N a_k s^k A\mathrm{e}^{st} = 0\),

leading to \(N\) roots \(s_i\).

Thus, \(y_h(t) = \sum_{i=1}^NA_i\mathrm{e}^{s_it}\) is the homogeneous solution, where \(A_i\) is undetermined.

Auxilliary conditions

To determine \(A_i\), typically we need \(N\) auxilliary conditions such as the value of

\[\frac{\mathrm{d}^k y_h(t)}{\mathrm{d}t^k} \bigg|_{t=t_0}, k=0,1,\ldots,N-1\]

If the system is a linear system, for zero input (the homogeneous equation), we get zero output. Thus the auxilliary conditions are all zero.

Casual LTI system

If the system is casual LTI system, it means that the system should satisfy the initial rest condition: if the input is zero prior to some time, then the output must be zero at least until the same time.

Linear Constant-Coefficient Difference Equations

Since it’s similar to the previous case, we will ignore some calculations here.

\[\sum_{k=0}^N a_k y[n-k] = \sum_{k=0}^M b_k x[n-k] \qquad\text{(1)}\]

Homogeneous Equation

\[\sum_{k=0}^N a_k y_h[n-k] = 0 \qquad\text{(2)}\]

The form of homogeneous solution

\[y[n] = \sum_{i=1}^N A_i z_i^n\]

Auxilliary conditions

Need \(N\) auxilliary conditions such as:

\[y[n] \bigg|_{n=n_0}, n=n_0, n_0=1, \ldots, n_0-N+1\]

Explicit solution to difference equation

\[y[n] = \frac{1}{a_0}(\sum_{k=0}^M b_k x[n-k] - \sum_{k=1}^N a_k y[n-k])\]

• Assume causality Initial rest until we start the upper equation.

Ways to save memory

By exchanging the order of the cascaded operations.