how to compute energy and power of discrete-time signal using symbolic calculations

I would like to get some parameters of continous and discrete-time signals i.a. the total energy and the mean power using symbolic calculations. I do not have problem to do this for a continous signal:

syms t
A = 10; %[1]
B = 5;
f = 10; %[Hz]
s = A*sin(2*pi*f*t) + B;

Et = int(s.^2,t,-inf,inf)
Pt = limit((int(s.^2,t,-t,t))/(2*t),t,inf)

but I do not have any idea how to do this for the signal which is discrete in time. I have tried, unsaccessfully, the symsym function

En = symsum(s.^2,t,-inf,inf)
Pn = limit((symsum(s.^2,t,-t,t))/(2*t),t,inf)

I expect the energy to be infinite and this my cause some problems, but the power should be a finite number.

Does anybody have an idea how to solve this problem?

Answer 1

[Swetha Polemoni]

Hi,

I have tried your code and power is finite for discrete signal also. Use "simplify" to get limit value instead of limit expression. Following code is for better understanding.

syms t 
A = 10; %[1] 
B = 5; 
f = 1; %[Hz] 
s = A*sin(2*pi*f*t) + B; 
En = simplify(symsum(s.^2,t,-inf,inf)) 
Pn =simplify( limit((symsum(s.^2,t,-t,t))/(2*t),t,inf))

Here Pn = 25. Analyzing it step by step

-> Consider the summation.

->Here Sin(200*pi*t) is always zero since it is being calculated at integer multiplies of 2*pi. So the summation results in 25*2t

->Now dividing the summation with 2t and applying limit t->infinite results in 25*2*t/2*t => Pn =25

Though Pn is not equal to Pt here, Pn=25 is correct. Pn and Pt are different, may be, because not every time discrete signal has same power/energy as continuos signal.

[BLP]

First of all I am not sure why do you say that in the case of sumation symsum(s.^2,t,-t,t)) is always zero. I agree that if t cause that we have a integer multiplies of 2*pi, the sumation of si function will be ze, but we do have a sum . I have used the symbolic calcolations to get the subtracts (see the code below)

syms t n
A = 10; %[1]
B = 5;
f = 1; %[Hz]
T = 1;
fp = 10;

s1 = A*sin(2*pi*f*t)
s5 = B;

sumS1sqr = symsum(s1.^2,t,-t,t)
sumS1sqr = symsum(2*5*s1,t,-t,t)
sumS5sqr = symsum(s5.^2,t,-t,t)

and I got the following outputs (see below)

s1 =

s5 = 5

sumS1sqr =

sum10S1 = 0

sumS5sqr =

Do you agree with this calculations?

Secondly, if the summation results in 25*2t, why the integration did not? I mean if limit((symsum(s1.^2,t,-t,t))/(2*t+ 1),t,inf) is always equal to 25, why limit((int(s1.^2,t,-t,t))/(2*t),t,inf) is not?

And finally, I have defined new signal s2 which is the result of sampling the s1 signal with the sampling frequency fp and then calcualted the energy and the power of the signal s2 and again I do not understand the outputs (see the code and the outputs below).

What am I doing wron?

syms t n
A = 10; %[1]
B = 5;
f = 1; %[Hz]
T = 1;
fp = 10;

s1 = A*sin(2*pi*f*t) + B;

Et = int(s1.^2,t,-inf,inf)
Pt = limit((int(s1.^2,t,-t,t))/(2*t),t,inf)

s2 = A*sin(2*pi*f*n/fp) + B;

En = simplify(symsum(s2.^2,n,-inf,inf))
Pn = simplify(limit((symsum(s2.^2,n,-n,n))/(2*n+ 1),n,inf))

Es1ct = ∞

Ps1ct = 75

Es1dt = ∞

Ps1dt = 25

Es2n =

Ps2n =

[Swetha Polemoni]

Hi,

• "sumS1sqr" and "sum10S1" is always zero for any t(only integers) value since the summation is discrete. So the total summation in discrete,

             S_discrete =sumS1sqr+sum10S1+sumS5sqr=50t+25

When Limit is applied to S_dis after dividing with 2*t, it would result in 25.

• In integration, it is not equal to zero because it is continuos and non integers(i.e., whole sine curve) are also considered. Consider the following code with basic example where limt value is 5

syms t
A = 10; 
B = 5;
f = 100; 
s = A*sin(2*pi*f*t) ;

%% WITH LIMIT %%
Using_Integration = limit((int(s.^2,t,-t,t)),t,5) % Output:Using_Integration= 500
Using_Summation = simplify(limit((symsum(s.^2,t,-t,t)),t,5))%Output: Using_Summation=0

%% WITHOUT LIMIT %%
Integration = ((int(s.^2,t,-10,10)))  %Output:Integration=700
Summation= simplify(symsum(s.^2,t,-10,10)) %Output: Summation=0

% Here though the same signal is considered and sum is calculated within same 
%limits, both outputs differ.

• I think it is not always possible to calculate energy or power in time domain. The same happend for the signal s2.

[BLP]

I am prety sure that the sumation of a sinus function of the lengthth qual to k*T, where T is the period of the sinus, will give zero - this may represent an area under the function, but in this case the sinus have both positive and negative valuesvalues which cancel each other totally. On the other hand, analogues sumation of suare of the sinus function cannot by equal to zero - because of the the square operation all values of the signal are positive.

sumS1sqr + sum10S1 + sumS5sqr

ans =

Of course using the limit for t->inf may give the final output equal to 25, but I do not understand why.

I expected some differences in the value pf the signal power for the continuous and discrete siganlas which will be due to the sampling, but when the power I get for the continuous signal is 75 and for the discrete ist 25 I expect an error in the code.

I thougt, that using the equations with t for the summation is wrong - I do not have any influence on the sampling interval. Therefore I have changed this and calculated new discrete signal and than I tried to calculate the energy and power but this did not help. Why - I do not know. Usnig fp and n instead of t I would like to check how the fp influence the power of the signal.

I have tried another way. I have defined a symbolic discrete signal and a vecotr o samples of discrete signal

syms n

A1 = 10; % [1]
B1 = 5;
f1 = 10; % [Hz] sinus frequency
fp2d = 100; % [Hz] sampling frequency
Tprob1d = 0.95; % [s] sampling time

ss2d = A1*sin(2*pi*f1*n/fp2d) + B1; % symbolic discrete signal

L = length(0:1/fp2d:Tprob1d);
s2d = subs(ss2d,1:L); % discrete signa as a vector

Than, I calculated power for both for a finite length equal one period

No2 = floor(fp2d/f1);
n1 = 1;
n2 = No2;

Pnss2dNo2 = double(limit((symsum(ss2d.^2,n,n1,n2))/(n2 - n1 + 1),n,inf))
Pns2dNo2 = double(1/(n2-n1+1)*sum(s2d(n1:n2).^2))

and the results are correct

Pnss2dNo2 = 75

Pns2dNo2 = 75

In this case, error caused by sampling equals zero.

Than I have changed the sampling frequency and repeted the calculations

syms n

A1 = 10; % [1]
B1 = 5;
f1 = 10; % [Hz] sinus frequency
fp2d = 20; % [Hz] sampling frequency
Tprob1d = 0.95; % [s] sampling time

ss2d = A1*sin(2*pi*f1*n/fp2d) + B1; % symbolic discrete signal

L = length(0:1/fp2d:Tprob1d);
s2d = subs(ss2d,1:L); % discrete signa as a vector

No2 = floor(fp2d/f1);
n1 = 1;
n2 = No2;

Pnss2dNo2 = double(limit((symsum(ss2d.^2,n,n1,n2))/(n2 - n1 + 1),n,inf))
Pns2dNo2 = double(1/(n2-n1+1)*sum(s2d(n1:n2).^2))

This time I got quite different results

Pnss2dNo2 = 25

Pns2dNo2 = 25

This in my opinion proves, that the problem is in the definition of the independetn variable (as t) which ends up in an erroneous sampling of the sinus function.

It does not chage the situation. Still, I do not have any idea why I canot get the proper value in a general case for infinite length of the signal ssd2 (se code and outputs below)?

Pnss2d = simplify(limit((symsum(ss2d.^2,n,-n,n))/(2*n + 1),n,inf))

Pnss2d =