LTI systems are ones which are both linear and time-invariant.
Impulse Response of LTI systems
The Discrete Case
Notice this operation is the convolution between the input and the impulse response.
The Continuous Case
After applying the LTI system to it,
Notice this operation is the convolution between the input and the impulse response.
Determining Properties of an LTI system
Because an LTI system is determined entirely by its impulse response, we can determine its properties from the impulse response.
Causality
Theorem 9
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Memory
Theorem 10
Stability
Theorem 11
so
Frequency Response and Transfer Functions
Definition 34
The frequency response of a system is the output when passed a purely oscillatory signal
If we pass a complex exponential into an LTI system, the output signal is the same signal but scaled. In otherwise, it is an eigenfunction of LTI systems.
The integral is a constant, and the original function is unchanged. The same analysis can be done in the discrete case.
We give these constant terms a special name called the transfer function.
Definition 35
Notice: The frequency response is the fourier transform of the impulse response! This means the Fourier Transform takes us from the impulse response of the system to the frequency response. There is no reason to limit ourselves to the Fourier Domain though
Definition 36
The transfer function is merely the Laplace Transform of the impulse response. In many ways, this can be more useful than the frequency response.
Stability of transfer functions
Recall that an LTI system is stable if the impulse response is absolutely integrable. We can determine this from the transfer function.
Theorem 12
Theorem 13
Bode Plots
Because transfer functions, and hence the frequency response, can be quite complex, we need a easy way to visualize how a system responds to different frequencies.
Definition 37
The log-log scale not only allows us to determine the behavior of Bode plots over a large range of frequencies, but they also let us easily figure out what the plot looks like because it converts the frequency response into piecewise linear components.
To see why, lets write our transfer function in polar form.
If we take the log of this, we get
Thus we can see at decades away from the poles and zeros, the magnitudes and the phases will have less of an effect. Let’s try constructing the Bode Plot for this transfer function.
Special LTI Systems
Linear Constant Coefficient Difference/Differential Equations
Definition 38
A linear constant coefficient difference equation is a system of one of the following forms
Theorem 14
Theorem 15
Systems of the form
are causal, FIR LTI systems and their impulse response is
Theorem 16
Given a constant coefficient difference/differential equation, the transfer function is
Proof. The Continuous Case
Taking the Laplace Transform,
The Discrete Case
Taking the Z Transform
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State Space Equations
When we have a LCCDE of the form
Important: This is valid in Discrete Time as well!
In general, state-space equations are useful because they allow us to find transfer functions of complex systems very easily.
Label the output of delay (discrete) or differentiation (continuous) blocks as the state variables.
Write the state equations using inputs and delays/derivatives. Express each as a weighted sum of states and inputs.
Use the formula above to find the transfer function.
Second Order Systems
Most of the time, higher order systems only have 2 dominant poles. Accordingly, they can be approximated by second order systems (i.e systems with two poles). One way to write the transfer function of this system is
Using common laplace transform pairs, this corresponds to
There are several key features of this step response:
Using the step response, we can calculate some of these values.
LTI systems are special systems because their output can be determined entirely the impulse response h[n].
We can think of the original signal x[n] in terms of the impulse function.
x[n]=x[0]δ[n]+x[1]δ[n−1]+...=∑k=−∞∞x[k]δ[n−k]
This signal will be transformed in some way to get the output y[n]. Since the LTI system applies a functional F and the LTI is linear and time-invariant,
This means as long as ∑k∣h[k] converges, y[n] will be bounded.
2. Assume ∑n∣h[n]∣ does not converge. Show that the system is unstable. Choose x[n]=sgn{h[−n]}
y[n]=∑kx[n−k]h[k]
y[0]=∑kx[−k]h[k]=∑k∣h[k]∣
And this is unbounded, so y[n] is unbounded. ◻
y(t)=∫−∞∞es(t−τ)h(τ)dτ=est∫−∞∞e−sτh(τ)
y[n]=∑k=−∞∞zn−kh[k]=zn∑k=−∞∞z−kh[k]
The frequency response of an LTI system H(jω) is how the system scales a pure tone of frequency ω
H(ω):=∫−∞∞h(τ)e−jωτdτ,H(ω):=∑k=−∞∞h[k]e−jωk
The transfer function of an LTI system H(s)is how the system responds to complex exponentials.
A causal continuous LTI system is stable iff all poles of H(s)have negative real parts.
The proof of this theorem stems from some facts about the Laplace Transform. If the system is causual, then the ROC is the half place demarcated by the right most pole. When this ROC includes the imaginary axis, the Fourier Transform is well defined, and this only happens when h(t) is absolutely integrable. Applying the same logic to the discrete case,
A causal discrete LTI system is stable iff all poles of H(z)lie within the unit circle.
This is because we know the ROC extends from the right-most pole for causal systems, and for the Fourier Transform to exist (making h[n] absolutely integrable), the ROC must contain the unit circle.
A Bode Plot is a straight-line approximation plot of ∣H(jω)∣ and ∠H(jω)on a log-log scale
Each r is the magnitude of a factor 1+ωnjω where ωn is either a root or a pole, and the zi,pk are the phases of each factor. By writing H(jω) this way, it is clear that
For Bode plots, we use the decibel scale, meaning we will multiply this value by 20 when constructing our plot. The exponential form of H(jω) tells us that
∠H(jω)=2π(Nz0−Np0)+(∑i=0nzi−∑k=0mpk)
Next, we should verify if we can approximate these equations as linear on a log-log scale. Take the example transfer function H(jω)=1+ωpjω1=rp1e−jθp.
if ω=ωpif ω=10ωpif ω=0.1ωpH(jω)=1+j1H(jω)=1+10j1H(jω)=1+0.1j1rp=2rp≈10rp≈1θp=4πθp≈2πθp≈0
For the magnitude plot, since there are no poles or zeros at ω=0, we draw a straight line until the pole kicks in at ω=ωp at which point the slope of the line will be -1. For the phase plot, we apply the same logic, except the pole kicks in at 10ωp (to see why, look above to see how at ω=ωp, the phase is −4π). We can apply this same logic for more complicated transfer functions too. Lets take
H(jω)=109(jω)(1+107jω)(1+109jω)
Notice we have a zero at 109, poles at 1,107, and 9 zeros at ω=0. With this information, we can see the plots will look like this:
The pole at 0 kicks in immediately, causing the decreasing magnitude and starting the phase at 2−π. The second pole at 107 will kick in next, followed by the zero at 109.
Discrete: ∑k=0Naky[n−k]=∑k=0Mbkx[n−k]
Continuous: ∑k=0Nakdtkdky=∑k=0Mbkdtkdkx
Systems described by a linear constant coefficient difference equation are causal LTI iff a0=0 and the system is initially at rest (y[n]=0 for n<n0 where n0 is the first instant x[n]=0)
Notice that if a1..an=0, then the system will have a finite impulse response because eventually the signal will die out. It turns out that all causal FIR systems can be written as a linear constant coefficient difference equation.
Notice that the poles of the transfer function are simply the eigenvalues of A. This is because if s is an eigenvalue of A, (sI−A)−1 is not invertible so the transfer function is undefined just like it is at the poles.
Write y[n] in terms of x[n] and the state variables
H(s)=s2+2ζωns+ωn2ωn2.
The parameter ζ is known as the damping ratio, and the parameter ωn is known as the natural frequency. This parameterization is useful because it gives us an insight into how the system will behave. First, notice that when ζ∈[0,1], we will get two complex poles. Suppose we want to find the impulse response of this system. Because the poles are complex, we can write the poles in the form
−ωncosθσ±jωnsinθωd
where ζ=cosθ. These new paramters allow us to rewrite our transfer function as
H(s)=(s+σ)2+ωd2ωn2
h(t)=ωdωn2e−σtsin(ωdt)u(t)
This is a damped sinusoid. Notice how ζ, which is related to σ controls the exponential and therefore the damping factor. If we find the step response of the second order system, we will get
y(t)=[1−e−σt(cosωdt+ωdσsinωdt)]u(t).
Rise Time (tr): Time to go from 10% to 90% of the steady state value
Peak Overshoot (Mp): steadypeak−steady
Peaking Time (tp): Time to peak overshoot
Settling Time (ts): Time after which step response stays within 1% if steady state.
