Use the symbolic toolbox of Matlab to calculate the energy of the signals x(t) and f(t). Verify theoretically. How is the energy of these two signals related?

I have two piecewise defined functions, x(t) and a reversed and shifted version of x(t) as shown below...I am supposed to use the symbolic toolbox in Matlab to calculate the energy of these two signals and explain how they are related. Not sure what the symbolic toolbox is...is it using syms function? Code for both signals is below.

 % code
x = @(t) zeros(size(t)) +(t >= 0 & t < 2).*((1/2)*exp(t)) + (t >= 2 & t < 4).*(2*t-8);
t = linspace(0,6);
plot(t,x(t))
axis([-1 8 -4 4])
xlabel('Time(t)')
ylabel('x(t)')
title('Function x(t)')

% code
x = @(t) zeros(size(t)) +(t >= 0 & t < 2).*((1/2)*exp(t)) + (t >= 2 & t < 4).*(2*t-8);
f = @(t) x(-t-1);
t = linspace(-7, -1);
plot(t,f(t))
axis([-8 1 -4 4])
xlabel('Time(t)')
ylabel('f(t)');
title('Left shift and time reversal of x(t)')

Answer 1

[Walter Roberson]

That code I showed before,

syms x(t) shifted_x(t)
x(t) = evalin(symengine, 'piecewise([0 <= t and t < 2, exp(t)/2], [2 <= t and t < 4, 2*t-8], [t < 0 or t >= 4, 0])');
shifted_x(t) = x(-t-1);

is the Symbolic Toolbox implementation of a piecewise function. This is something that I would not expect very many people to know.

Possibly you are expected to implement in terms of heaviside functions and dirac delta

syms F(t)
F(t) = -exp(t)/2 * heaviside(-2 + t) + exp(t)/2 * heaviside(t) - (2 * t - 8) * heaviside(-4 + t) + (2 * t  -8 ) * heaviside(-2 + t) + dirac(t)*(1/2) + dirac(t-2)*(-4)

The value of heaviside at the point where the parameter is 0 is something that can be adjusted with sympref() from R2015a onwards. The issue here is that heaviside(t) is 0 for t < 0 and is 1 for t > 0, but what is it at exactly t = 0 ? The dirac() is there to give explicit values at the boundary conditions

According to https://www.mathworks.com/matlabcentral/newsreader/view_thread/320985

the energy of a signal F is var(F)

In turn, var(F(t)) is E(F(t)^2)-E(F(t))^2 where E standards for expected value. The expected value of F(t) is int(t*F(t), t, -inf, inf)

You can now put those formulas together with heaviside version of your piecewise functions to come up with a value for their energies. F(t) and F(-1-t) are the formula...

[Aaron Connell]

hmm. I am still a bit confused. Say I want to use the heaviside, when I try var(F(t)) I get Nan...

[Walter Roberson]

Do not use var() for symbolic expressions, use the expectation based formula I showed, it defines variance for continuous functions. var() is only for discrete distributions.

[Aaron Connell]

it returns NaN

[Aaron Connell]

    % code
syms F(t)
F(t) = -exp(t)/2 * heaviside(-2 + t) + exp(t)/2 * heaviside(t) - (2 * t - 8) * heaviside(-4 + t) + (2 * t  -8 ) * heaviside(-2 + t) + dirac(t)*(1/2) + dirac(t-2)*(-4)
energy=int(t*F(t), t, -inf, inf)
  end

[Aaron Connell]

Then I tried

 %
syms F(t)
F(t) = -exp(t)/2 * heaviside(-2 + t) + exp(t)/2 * heaviside(t) - (2 * t - 8) * heaviside(-4 + t) + (2 * t  -8 ) * heaviside(-2 + t) + dirac(t)*(1/2) + dirac(t-2)*(-4)
E=int(t*F(t), t, -inf, inf)
Energy=E(F(t)^2)-E(F(t)^2)
  end

[Walter Roberson]

Sorry I need to sleep now

[Karan Gill]

Piecewise is available starting 16b: http://www.mathworks.com/help/symbolic/piecewise.html

Answer 2

[Paul]

syms x(t)
assume(t,'real');
x(t) = exp(t)/2*rectangularPulse(0,2,t) + (2*t - 8)*rectangularPulse(2,4,t)

Energy in real-valued x(t):

Ex = int(x(t)^2,t,-inf,inf)
disp(char(Ex))

exp(4)/8 + 253/24

x1(t) = x(-t-1);
Ex1 = int(x1(t)^2,t,-inf,inf)
disp(char(Ex1))

exp(4)/8 + 253/24