## Discrete Fourier Transform

In [1] a verification of the discrete Fourier transform pairs was performed. A much different looking discrete Fourier transform pair is given in [2] \$ A.4. This transform pair samples the points at what are called the Nykvist time instants given by

\label{eqn:discreteFourier:20}
t_k = \frac{T k}{2 N + 1}, \qquad k \in [-N, \cdots N]

Note that the endpoints of these sampling points are not $$\pm T/2$$, but are instead at

\label{eqn:discreteFourier:40}
\pm \frac{T}{2} \inv{1 + 1/N},

which are slightly within the interior of the $$[-T/2, T/2]$$ range of interest. The reason for this slightly odd seeming selection of sampling times becomes clear if one calculate the inversion relations.

Given a periodic ($$\omega_0 T = 2 \pi$$) bandwith limited signal evaluated only at the Nykvist times $$t_k$$,

\label{eqn:discreteFourier:60}
\boxed{
x(t_k) = \sum_{n = -N}^N X_n e^{ j n \omega_0 t_k},
}

assume that an inversion relation can be found. To find $$X_n$$ evaluate the sum

\label{eqn:discreteFourier:80}
\begin{aligned}
&\sum_{k = -N}^N x(t_k) e^{-j m \omega_0 t_k} \\
\sum_{k = -N}^N
\lr{
\sum_{n = -N}^N X_n e^{ j n \omega_0 t_k}
}
e^{-j m \omega_0 t_k} \\
\sum_{n = -N}^N X_n
\sum_{k = -N}^N
e^{ j (n -m )\omega_0 t_k}
\end{aligned}

This interior sum has the value $$2 N + 1$$ when $$n = m$$. For $$n \ne m$$, and
$$a = e^{j (n -m ) \frac{2 \pi}{2 N + 1}}$$, this is

\label{eqn:discreteFourier:100}
\begin{aligned}
\sum_{k = -N}^N
e^{ j (n -m )\omega_0 t_k}
&=
\sum_{k = -N}^N
e^{ j (n -m )\omega_0 \frac{T k}{2 N + 1}} \\
&=
\sum_{k = -N}^N a^k \\
&=
a^{-N} \sum_{k = -N}^N a^{k+ N} \\
&=
a^{-N} \sum_{r = 0}^{2 N} a^{r} \\
&=
a^{-N} \frac{a^{2 N + 1} – 1}{a – 1}.
\end{aligned}

Since $$a^{2 N + 1} = e^{2 \pi j (n – m)} = 1$$, this sum is zero when $$n \ne m$$. This means that

\label{eqn:discreteFourier:120}
\sum_{k = -N}^N
e^{ j (n -m )\omega_0 t_k} = (2 N + 1) \delta_{n,m},

which provides the desired Fourier inversion relation

\label{eqn:discreteFourier:140}
\boxed{
X_m = \inv{2 N + 1} \sum_{k = -N}^N x(t_k) e^{-j m \omega_0 t_k}.
}

# References

[1] Peeter Joot. Condensed matter physics., appendix: Discrete Fourier transform. 2013. URL https://peeterjoot.com/archives/math2013/phy487.pdf. [Online; accessed 02-December-2014].

[2] Giannini and Leuzzi Nonlinear Microwave Circuit Design. Wiley Online Library, 2004.

## Laplace transform refresher

November 27, 2014 ece1254 , , , ,

Laplace transforms were used to solve the MNA equations for time dependent systems, and to find the moments used to in MOR.

For the record, the Laplace transform is defined as:

\label{eqn:laplaceTransformVec:20}
\boxed{
\LL( f(t) ) =
\int_0^\infty e^{-s t} f(t) dt.
}

The only Laplace transform pair used in the lectures is that of the first derivative

\label{eqn:laplaceTransformVec:40}
\begin{aligned}
\LL(f'(t)) &=
\int_0^\infty e^{-s t} \ddt{f(t)} dt \\
&=
\evalrange{e^{-s t} f(t)}{0}{\infty} – (-s) \int_0^\infty e^{-s t} f(t) dt \\
&=
-f(0) + s \LL(f(t)).
\end{aligned}

Here it is loosely assumed that the real part of $$s$$ is positive, and that $$f(t)$$ is “well defined” enough that $$e^{-s \infty } f(\infty) \rightarrow 0$$.

Where used in the lectures, the laplace transforms were of vectors such as the matrix vector product $$\LL(\BG \Bx(t))$$. Because such a product is linear, observe that it can be expressed as the original matrix times a vector of Laplace transforms

\label{eqn:laplaceTransformVec:60}
\begin{aligned}
\LL( \BG \Bx(t) )
&=
\LL {\begin{bmatrix}
G_{i k} x_k(t)
\end{bmatrix}}_i \\
&=
{\begin{bmatrix}
G_{i k} \LL x_k(t)
\end{bmatrix}}_i \\
&=
\BG
{\begin{bmatrix}
\LL x_i(t)
\end{bmatrix}}_i.
\end{aligned}

The following notation was used in the lectures for such a vector of Laplace transforms

\label{eqn:laplaceTransformVec:80}
\BX(s) = \LL \Bx(t) =
{\begin{bmatrix}
\LL x_i(t)
\end{bmatrix}}_i.

## ECE1254H Modeling of Multiphysics Systems. Lecture 19: Model order reduction (cont).. Taught by Prof. Piero Triverio

November 26, 2014 ece1254 , , ,

## Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

## Model order reduction (cont).

An approximation of the following system is sought

\label{eqn:multiphysicsL19:20}
\BG \Bx(t) + C \dot{\Bx}(t) = B \Bu(t)

\label{eqn:multiphysicsL19:40}
\By(t) = \BL^\T \Bx(t).

The strategy is to attempt to find a $$N \times q$$ projector $$\BV$$ of the form

\label{eqn:multiphysicsL19:60}
\BV =
\begin{bmatrix}
\Bv_1 & \Bv_2 & \cdots & \Bv_q
\end{bmatrix}

so that the solution of the constrained q-variable state vector $$\Bx_q$$ is sought after letting

\label{eqn:multiphysicsL19:80}
\Bx(t) = \BV \Bx_q(t).

## Moment matching

\label{eqn:multiphysicsL19:100}
\begin{aligned}
\BF(s)
&= \lr{ \BG + s \BC }^{-1} \BB \\
&= \BM_0 + \BM_1 s + \BM_2 s^2 + \cdots + \BM_{q-1} s^{q-1} + M_q s^q + \cdots
\end{aligned}

The reduced model is created such that

\label{eqn:multiphysicsProblemSet3b:120}
\BF_q(s)
=
\BM_0 + \BM_1 s + \BM_2 s^2 + \cdots + \BM_{q-1} s^{q-1} + \tilde{\BM}_q s^q.

Form an $$N \times q$$ projection matrix

\label{eqn:multiphysicsL19:140}
\BV_q \equiv
\begin{bmatrix}
\BM_0 & \BM_1 & \cdots & \BM_{q-1}
\end{bmatrix}

With the substitution of fig. 1, becomes

This is a system of $$N$$ equations, in $$q$$ unknowns. A set of moments from the frequency domain have been used to project the time domain system. This relies on the following unproved theorem (references to come)

### Theorem

If $$\text{span}\{ \Bv_q \} = \text{span} \{ \BM_0, \BM_1, \cdots, \BM_{q-1} \}$$, then the reduced model will match the first $$q$$ moments.

Left multiplication by $$\BV_q^\T$$ yields fig. 2.

This is now a system of $$q$$ equations in $$q$$ unknowns.

With

\label{eqn:multiphysicsL19:160}
\BG_q = \BV_q^\T \BG \BV_q

\label{eqn:multiphysicsL19:180}
\BC_q = \BV_q^\T \BC \BV_q

\label{eqn:multiphysicsL19:200}
\BB_q = \BV_q^\T \BB

\label{eqn:multiphysicsL19:220}
\BL_q^\T = \BL^\T \BV_q

the system is reduced to

\label{eqn:multiphysicsL19:240}
\BG_q \Bx_q(t) + \BC_q \dot{\Bx}_q(t) = \BB_q \Bu(t).

\label{eqn:multiphysicsL19:260}
\By(t) \approx \BL_q^\T \Bx_q(t)

### Moments calculation

Using
\label{eqn:multiphysicsL19:300}
\lr{ \BG + s \BC }^{-1} \BB = \BM_0 + \BM_1 s + \BM_2 s^2 + \cdots

thus
\label{eqn:multiphysicsL19:320}
\begin{aligned}
\BB &=
\lr{ \BG + s \BC }
\BM_0 +
\lr{ \BG + s \BC }
\BM_1 s +
\lr{ \BG + s \BC }
\BM_2 s^2 + \cdots \\
&=
\BG \BM_0
+ s \lr{ C \BM_0 + \BG \BM_1 }
+ s^2 \lr{ C \BM_1 + \BG \BM_2 }
+ \cdots
\end{aligned}

Since $$\BB$$ is a zeroth order matrix, setting all the coefficients of $$s$$ equal to zero provides a method to solve for the moments

\label{eqn:multiphysicsL19:340}
\begin{aligned}
\BB &= \BG \BM_0 \\
-\BC \BM_0 &= \BG \BM_1 \\
-\BC \BM_1 &= \BG \BM_2 \\
\end{aligned}

The moment $$\BM_0$$ can be found with LU of $$\BG$$, plus the forward and backward substitutions. Proceeding recursively, using the already computed LU factorization, each subsequent moment calculation requires only one pair of forward and backward substitutions.

Numerically, each moment has the exact value

\label{eqn:multiphysicsL19:360}
\BM_q = \lr{- \BG^{-1} \BC }^q \BM_0.

As $$q \rightarrow \infty$$, this goes to some limit, say $$\Bw$$. The value $$\Bw$$ is related to the largest eigenvalue of $$-\BG^{-1} \BC$$. Incidentally, this can be used to find the largest eigenvalue of $$-\BG^{-1} \BC$$.

The largest eigenvalue of this matrix will dominate these factors, and can cause some numerical trouble. For this reason it is desirable to avoid such explicit moment determination, instead using implicit methods.

The key is to utilize the theorem above, and look instead for an alternate basis $$\{ \Bv_q \}$$ that also spans $$\{ \BM_0, \BM_1, \cdots, \BM_q \}$$.

### Space generate by the moments

Write

\label{eqn:multiphysicsL19:380}
\BM_q = \BA^q \BR,

where in this case

\label{eqn:multiphysicsL19:400}
\begin{aligned}
\BA &= – \BG^{-1} \BC \\
\BR &= \BM_0 = -\BG \BB
\end{aligned}

The span of interest is

\label{eqn:multiphysicsL19:420}
\text{span} \{ \BR, \BA \BR, \BA^2 \BR, \cdots, \BA^{q-1} \BR \}.

Such a sequence is called a Krylov subspace. One method to compute such a basis, the Arnoldi process, relies on Gram-Schmidt orthonormalization methods:

Some numerical examples and plots on the class slides.

## Stability of discretized linear differential equations

November 17, 2014 ece1254 , , , ,

In class today was a highlight of stability methods for linear multistep methods. To motivate the methods used, it is helpful to take a step back and review stability concepts for LDE systems.

By way of example, consider a second order LDE homogeneous system defined by

\label{eqn:stabilityLDEandDiscreteTime:20}
\frac{d^2 x}{dt^2} + 3 \frac{dx}{dt} + 2 = 0.

Such a system can be solved by assuming an exponential solution, say

\label{eqn:stabilityLDEandDiscreteTime:40}
x(t) = e^{s t}.

Substitution gives

\label{eqn:stabilityLDEandDiscreteTime:60}
e^{st} \lr{ s^2 + 3 s + 2 } = 0,

The polynomial part of this equation, the characteristic equation has roots $$s = -2, -1$$.

The general solution of \ref{eqn:stabilityLDEandDiscreteTime:20} is formed by a superposition of solutions for each value of $$s$$

\label{eqn:stabilityLDEandDiscreteTime:80}
x(t) = a e^{-2 t} + b e^{-t}.

Independent of any selection of the superposition constants $$a, b$$, this function will not blow up as $$t \rightarrow \infty$$.

This “stability” is due to the fact that both of the characteristic equation roots lie in the left hand Argand plane.

Now consider a discretized form of this LDE. This will have the form

\label{eqn:stabilityLDEandDiscreteTime:100}
\begin{aligned}
0 &=
\inv{\lr{\Delta t}^2}
\lr{ x_{n+2} – 2 x_{n-1} + x_n } + \frac{3}{\Delta t} \lr{ x_{n+1} – x_n } + 2
x_n \\
&=
x_{n+2} \lr{
\inv{\lr{\Delta t}^2}
}
+
x_{n+1} \lr{
\frac{3}{\Delta t}
-\frac{2}{\lr{\Delta t}^2}
}
+
x_{n} \lr{
\frac{1}{\lr{\Delta t}^2}
-\frac{3}{\Delta t}
+ 2
},
\end{aligned}

or

\label{eqn:stabilityLDEandDiscreteTime:220}
0
=
x_{n+2}
+
x_{n+1} \lr{
3 \Delta t – 2
}
+
x_{n} \lr{
1 – 3 \Delta t + 2 \lr{ \Delta t}^2
}.

Note that after discretization, each subsequent index corresponds to a time shift. Also observe that the coefficients of this discretized equation are dependent on the discretization interval size $$\Delta t$$. If the specifics of those coefficients are ignored, a general form with the following structure can be observed

\label{eqn:stabilityLDEandDiscreteTime:120}
0 =
x_{n+2} \gamma_0
+
x_{n+1} \gamma_1
+
x_{n} \gamma_2.

It turns out that, much like the LDE solution by characteristic polynomial, it is possible to attack this problem by assuming a solution of the form

\label{eqn:stabilityLDEandDiscreteTime:140}
x_n = C z^n.

A time shift index change $$x_n \rightarrow x_{n+1}$$ results in a power adjustment in this assumed solution. This substitution applied to \ref{eqn:stabilityLDEandDiscreteTime:120} yields

\label{eqn:stabilityLDEandDiscreteTime:160}
0 =
C z^n
\lr{
z^{2} \gamma_0
+
z \gamma_1
+
1 \gamma_2
},

Suppose that this polynomial has roots $$z \in \{z_1, z_2\}$$. A superposition, such as

\label{eqn:stabilityLDEandDiscreteTime:180}
x_n = a z_1^n + b z_2^n,

will also be a solution since insertion of this into the RHS of \ref{eqn:stabilityLDEandDiscreteTime:120} yields

\label{eqn:stabilityLDEandDiscreteTime:200}
a z_1^n
\lr{
z_1^{2} \gamma_0
+
z_1 \gamma_1
+
\gamma_2
}
+
b
z_2^n
\lr{
z_2^{2} \gamma_0
+
z_2 \gamma_1
+
\gamma_2
}
=
a z_1^n \times 0
+b z_2^n \times 0.

The zero equality follows since $$z_1, z_2$$ are both roots of the characteristic equation for this discretized LDE.
In the discrete $$z$$ domain stability requires that the roots satisfy the bound $$\Abs{z} < 1$$, a different stability criteria than in the continuous domain. In fact, there is no a-priori guarantee that stability in the continuous domain will imply stability in the discretized domain. Let's plot those z-domain roots for this example LDE, using $$\Delta t \in \{ 1/2, 1, 2 \}$$. The respective characteristic polynomials are $$\label{eqn:stabilityLDEandDiscreteTime:260} 0 = z^2 - \inv{2} z = z \lr{ z - \inv{2} }$$ $$\label{eqn:stabilityLDEandDiscreteTime:240} 0 = z^2 + z = z\lr{ z + 1 }$$ $$\label{eqn:stabilityLDEandDiscreteTime:280} 0 = z^2 + 4 z + 3 = (z + 3)(z + 1).$$ These have respective roots $$\label{eqn:stabilityLDEandDiscreteTime:300} z = 0, \inv{2}$$ $$\label{eqn:stabilityLDEandDiscreteTime:320} z = 0, -1$$ $$\label{eqn:stabilityLDEandDiscreteTime:340} z = -1, -3$$ Only the first discretization of these three yields stable solutions in the z domain, although it appears that $$\Delta t = 1$$ is right on the boundary.

## ECE1254H Modeling of Multiphysics Systems. Lecture 16: LMS systems and stability. Taught by Prof. Piero Triverio

November 17, 2014 ece1254 , , ,

## Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

## Residual for LMS methods

### Mostly on slides:

12_ODS.pdf

Residual is illustrated in fig. 1, assuming that the iterative method was accurate until $$t_{n}$$

fig. 1. Residual illustrated

### Summary

• [FE]: $$R_{n+1} \sim \lr{ \Delta t}^2$$. This is of order $$p = 1$$.
• [BE]: $$R_{n+1} \sim \lr{ \Delta t}^2$$. This is of order $$p = 1$$.
• [TR]: $$R_{n+1} \sim \lr{ \Delta t}^3$$. This is of order $$p = 2$$.
• [BESTE]: $$R_{n+1} \sim \lr{ \Delta t}^4$$. This is of order $$p = 3$$.

## Global error estimate

Suppose $$t \in [0, 1] s$$, with $$N = 1/{\Delta t}$$ intervals. For a method with local error of order $$R_{n+1} \sim \lr{ \Delta t}^2$$ the global error is approximately $$N R_{n+1} \sim \Delta t$$.

## Stability

Recall that a linear multistep method (LMS) was a system of the form

\label{eqn:multiphysicsL16:20}
\sum_{j=-1}^{k-1} \alpha_j x_{n-j} = \Delta t \sum_{j=-1}^{k-1} \beta_j f( x_{n-j}, t_{n-j} )

Consider a one dimensional test problem

\label{eqn:multiphysicsL16:40}
\dot{x}(t) = \lambda x(t)

where as in fig. 2, $$\Re(\lambda) < 0$$ is assumed to ensure stability.

fig. 2. Stable system

Linear stability theory can be thought of as asking the question: “Is the solution of \ref{eqn:multiphysicsL16:40} computed by my LMS method also stable?”

Application of \ref{eqn:multiphysicsL16:20} to \ref{eqn:multiphysicsL16:40} gives

\label{eqn:multiphysicsL16:60}
\sum_{j=-1}^{k-1} \alpha_j x_{n-j} = \Delta t \sum_{j=-1}^{k-1} \beta_j \lambda x_{n-j},

or
\label{eqn:multiphysicsL16:80}
\sum_{j=-1}^{k-1} \lr{ \alpha_j – \Delta \beta_j \lambda }
x_{n-j} = 0.

With

\label{eqn:multiphysicsL16:100}
\gamma_j = \alpha_j – \Delta \beta_j \lambda,

this expands to
\label{eqn:multiphysicsL16:120}
\gamma_{-1} x_{n+1}
+
\gamma_{0} x_{n}
+
\gamma_{1} x_{n-1}
+
\cdots
+
\gamma_{k-1} x_{n-k} .

This can be seen as a

• discrete time system
• FIR filter

The numerical solution $$x_n$$ will be stable if \ref{eqn:multiphysicsL16:120} is stable.

A characteristic equation associated with \ref{eqn:multiphysicsL16:120} can be defined as

\label{eqn:multiphysicsL16:140}
\gamma_{-1} z^k
+
\gamma_{0} z^{k-1}
+
\gamma_{1} z^{k-2}
+
\cdots
+
\gamma_{k-1} = 0.

This is a polynomial with roots $$z_n$$ (poles). This is stable if the poles satisfy $$\Abs{z_n} < 1$$, as illustrated in fig. 3

Stability

Observe that the $$\gamma’s$$ are dependent on $$\Delta t$$.

FIXME: There’s a lot of handwaving here that could use more strict justification. Check if the text covers this in more detail.

### Example: Forward Euler stability

For $$k = 1$$ step.

\label{eqn:multiphysicsL16:180}
x_{n+1} – x_n = \Delta t f( x_n, t_n ),

the coefficients are $$\alpha_{-1} = 1, \alpha_0 = -1, \beta_{-1} = 0, \beta_0 =1$$. For the simple function above

\label{eqn:multiphysicsL16:200}
\gamma_{-1} = \alpha_{-1} – \Delta t \lambda \beta_{-1} = 1

\label{eqn:multiphysicsL16:220}
\gamma_{0} = \alpha_{0} – \Delta t \lambda \beta_{0} = -1 – \Delta t \lambda.

The stability polynomial is

\label{eqn:multiphysicsL16:240}
1 z + \lr{ -1 – \Delta t \lambda} = 0,

or

\label{eqn:multiphysicsL16:260}
\boxed{
z = 1 + \delta t \lambda.
}

This is the root, or pole.

For stability we must have

\label{eqn:multiphysicsL16:280}
\Abs{ 1 + \Delta t \lambda } < 1,

or
\label{eqn:multiphysicsL16:300}
\Abs{ \lambda – \lr{ -\inv{\Delta t} } } < \inv{\Delta t},

This inequality is illustrated roughly in fig. 4.

fig. 4. Stability region of FE

All poles of my system must be inside the stability region in order to get stable $$\gamma$$.