How to implement Complex Exponential Function Using Matlab ?

Please

How can I implement the Complex Exponential Function mentioned in this equation:

X(N) =exp^(j*w1*N) + exp^(j*w2*N)

Suppose:

F =40KHZ

N=Range 1 to 20

w1: is the first angular frequency

w1=2*pi*F

w2: is the second angular frequency

w2=2*pi*F

Finally,

display the values are saved in X(N) in command window.

We apply fast fourier transform (FFT) on X(N) Then we plot the result.

Answer 1

[Star Strider]

I am not certain what you are doing or what ‘w1’ and ‘w2’ are, so using my best guess —

format longE
F = 40E+3;
N = (0:20).';
omega1 = 2*pi*F;                                    % Guessing: 'W1'
omega2 = 4*pi*F;                                    % Guessing: 'w2'

X = exp(1j*omega1*N) + exp(1j*omega2*N)

X = 
      2.000000000000000e+00 + 0.000000000000000e+00i
      2.000000000000000e+00 - 1.165637649496438e-11i
      2.000000000000000e+00 - 2.331275298992875e-11i
      2.000000000000000e+00 - 3.496912948489312e-11i
      2.000000000000000e+00 - 4.662550597985751e-11i
      2.000000000000000e+00 + 2.909640830059826e-10i
      2.000000000000000e+00 - 6.993825896978625e-11i
      2.000000000000000e+00 - 4.308406009455551e-10i
      2.000000000000000e+00 - 9.325101195971502e-11i
      2.000000000000000e+00 + 2.443385770261251e-10i
      2.000000000000000e+00 + 5.819281660119652e-10i
      2.000000000000000e+00 - 4.774661069254125e-10i
      2.000000000000000e+00 - 1.398765179395725e-10i
      2.000000000000000e+00 + 1.977130710462676e-10i
      2.000000000000000e+00 - 8.616812018911102e-10i
      2.000000000000000e+00 - 5.240916129052701e-10i
      2.000000000000000e+00 - 1.865020239194300e-10i
      2.000000000000000e+00 - 1.245896296856808e-09i
      2.000000000000000e+00 + 4.886771540522501e-10i
      2.000000000000000e+00 - 5.707171188851276e-10i
      2.000000000000000e+00 + 1.163856332023930e-09i

.

[Muhammad Salem]

Star Strider

I'm sorry, I updated the question, please reconsider it :

w1: is the first angular frequency

w1=2*pi*F

w2: is the second angular frequency

w2=2*pi*F

[Image Analyst]

I'm sure you made the obvious change:

omega2 = 2*pi*F; 

so what went wrong? Why are you still asking? I can only assume it's not working for you. Be explicit about why Star's code with your new change of omega2 value is not working.

[Muhammad Salem]

Image Analyst

Yes, I am asking because I have a problem with the results and I do not know what the solution is.

Here are the results:

2.0000 + 0.0000i

2.0000 - 0.0000i

2.0000 - 0.0000i

2.0000 - 0.0000i

2.0000 - 0.0000i

2.0000 + 0.0000i

2.0000 - 0.0000i

2.0000 - 0.0000i

2.0000 - 0.0000i

2.0000 + 0.0000i

2.0000 + 0.0000i

2.0000 - 0.0000i

2.0000 - 0.0000i

2.0000 + 0.0000i

2.0000 - 0.0000i

2.0000 - 0.0000i

2.0000 - 0.0000i

2.0000 - 0.0000i

2.0000 + 0.0000i

2.0000 - 0.0000i

2.0000 + 0.0000i

[Star Strider]

O.K., so ‘w1’ and ‘w2’ are actually the same thing.

My code then becomes:

format longE
F = 40E+3;
N = (0:20).';
omega1 = 2*pi*F;                                    % Guessing: 'W1'
omega2 = 2*pi*F;                                    % Guessing: 'w2'

X = exp(1j*omega1*N) + exp(1j*omega2*N)

X = 
      2.000000000000000e+00 + 0.000000000000000e+00i
      2.000000000000000e+00 - 7.770917663309584e-12i
      2.000000000000000e+00 - 1.554183532661917e-11i
      2.000000000000000e+00 - 2.331275298992875e-11i
      2.000000000000000e+00 - 3.108367065323834e-11i
      2.000000000000000e+00 + 1.939760553373217e-10i
      2.000000000000000e+00 - 4.662550597985750e-11i
      2.000000000000000e+00 - 2.872270672970367e-10i
      2.000000000000000e+00 - 6.216734130647667e-11i
      2.000000000000000e+00 + 1.628923846840834e-10i
      2.000000000000000e+00 + 3.879521106746434e-10i
      2.000000000000000e+00 - 3.183107379502750e-10i
      2.000000000000000e+00 - 9.325101195971500e-11i
      2.000000000000000e+00 + 1.318087140308451e-10i
      2.000000000000000e+00 - 5.744541345940735e-10i
      2.000000000000000e+00 - 3.493944086035134e-10i
      2.000000000000000e+00 - 1.243346826129533e-10i
      2.000000000000000e+00 - 8.305975312378718e-10i
      2.000000000000000e+00 + 3.257847693681668e-10i
      2.000000000000000e+00 - 3.804780792567517e-10i
      2.000000000000000e+00 + 7.759042213492868e-10i

format shortE
X

X = 
   2.0000e+00 + 0.0000e+00i
   2.0000e+00 - 7.7709e-12i
   2.0000e+00 - 1.5542e-11i
   2.0000e+00 - 2.3313e-11i
   2.0000e+00 - 3.1084e-11i
   2.0000e+00 + 1.9398e-10i
   2.0000e+00 - 4.6626e-11i
   2.0000e+00 - 2.8723e-10i
   2.0000e+00 - 6.2167e-11i
   2.0000e+00 + 1.6289e-10i
   2.0000e+00 + 3.8795e-10i
   2.0000e+00 - 3.1831e-10i
   2.0000e+00 - 9.3251e-11i
   2.0000e+00 + 1.3181e-10i
   2.0000e+00 - 5.7445e-10i
   2.0000e+00 - 3.4939e-10i
   2.0000e+00 - 1.2433e-10i
   2.0000e+00 - 8.3060e-10i
   2.0000e+00 + 3.2578e-10i
   2.0000e+00 - 3.8048e-10i
   2.0000e+00 + 7.7590e-10i

whos('X')

  Name       Size            Bytes  Class     Attributes

  X         21x1               336  double    complex   

The imaginary parts are very close to zero, so using a more restricted format option, they will appear as zero. Using extended precision, this is readily apparent, and with greater precision or exponential notation (or both) this becomes clear.

EDIT — (3 Nov 2023 at 20:43)

My code does exactly what you asked, both in the original and in the edit. The ‘X’ result is a (21x1) column vector, showing both the real and imaginary parts. The size is (21x1) because ‘N’ includes zero, for a total of 21 elements. If you want it to be exactly (20x1) use the linspace function —

format longE
F = 40E+3;
N = linspace(0, 20, 20).'

N = 20×1

0 1.052631578947368e+00 2.105263157894737e+00 3.157894736842105e+00 4.210526315789473e+00 5.263157894736842e+00 6.315789473684211e+00 7.368421052631579e+00 8.421052631578947e+00 9.473684210526315e+00

omega1 = 2*pi*F;
omega2 = 2*pi*F;

X = exp(1j*omega1*N) + exp(1j*omega2*N)

X = 
      2.000000000000000e+00 + 0.000000000000000e+00i
     -1.651586909120879e-01 + 1.993168986016039e+00i
     -1.972722606816205e+00 - 3.291891804969826e-01i
      4.909709741865338e-01 - 1.938800531902734e+00i
      1.891634483443725e+00 + 6.493989382856979e-01i
     -8.033908495827032e-01 + 1.831546653188715e+00i
     -1.758947502506326e+00 - 9.518947859016555e-01i
      1.093896316053294e+00 - 1.674332956650209e+00i
      1.578281018953408e+00 + 1.228425425172970e+00i
     -1.354563142692437e+00 + 1.471447821860900e+00i
     -1.354563142806782e+00 - 1.471447821755638e+00i
      1.578281018857948e+00 - 1.228425425295617e+00i
      1.093896316573241e+00 + 1.674332956310511e+00i
     -1.758947502432355e+00 + 9.518947860383419e-01i
     -8.033908497250313e-01 - 1.831546653126284e+00i
      1.891634483393261e+00 - 6.493989384326954e-01i
      4.909709747886086e-01 + 1.938800531750268e+00i
     -1.972722606790624e+00 + 3.291891806502813e-01i
     -1.651586924591879e-01 - 1.993168985887843e+00i
      2.000000000000000e+00 + 7.759042213492868e-10i

format short
X

X = 
   2.0000 + 0.0000i
  -0.1652 + 1.9932i
  -1.9727 - 0.3292i
   0.4910 - 1.9388i
   1.8916 + 0.6494i
  -0.8034 + 1.8315i
  -1.7589 - 0.9519i
   1.0939 - 1.6743i
   1.5783 + 1.2284i
  -1.3546 + 1.4714i
  -1.3546 - 1.4714i
   1.5783 - 1.2284i
   1.0939 + 1.6743i
  -1.7589 + 0.9519i
  -0.8034 - 1.8315i
   1.8916 - 0.6494i
   0.4910 + 1.9388i
  -1.9727 + 0.3292i
  -0.1652 - 1.9932i
   2.0000 + 0.0000i

.

[Muhammad Salem]

Star Strider

Can you do FOR LOOP instead of using linspace ?

[Star Strider]

The linspace function just creates the vector with 20 evenly-spaced elements between 0 and 20 (a opposed to the colon operator that produces 21 elements). A for loop would not change that from the vectorised version. It would simply be a bit less efficient.

It is possible to do that, however a for loop (without using linspace to create the ‘N’ vector) would produce the same as the colon operator. It would not replace the linspace call.

[Muhammad Salem]

Star Strider

Thanks for your answer...

But I am required to implement it using For Loop. Can you implement it and tell me the results?

[Walter Roberson]

Nvalues = linspace(0, 20, 20).';
num_N = length(Nvalues);
X = zeros(num_N, 1);

for Nidx = 1 : num_N
    N = Nvalues(Nidx);
    now do your calculation for this N
    X(Nidx) = result for this N;
end

[Muhammad Salem]

Walter Roberson

Why did you use the linspace command?

I don't need it because the for loop (Nidx) will be in place N.

[Walter Roberson]

@Muhammad Salem

Your requirements from the initial question were:

N=Range 0 to 20
Finally, the values are saved in a vector 20 * 1

So you need the initial value to be 0, the final value to be 20, and for there to be 20 values total.

In order to do that as a for loop without using the linspace structure I show here, you would need

for N = 0:20/19:20

which will get you N values such as 0 1.05263157894737 2.10526315789474 but the 20'th will be 20.

Now that you are looping incrementing by 20/19 at a time... how will you know what index to store the current result at?

[Star Strider]

@Muhammad Salem — As always, my pleasure!

@Walter Roberson — Thank you!

[Muhammad Salem]

@Star Strider

@Walter Roberson

I am really sorry

The question has been updated. Perhaps the question contains some errors or may not be clear.

Please give me the answer again

[Star Strider]

That simply requires minor tweaks to the existing code.

See the documentation on the fft function to perform the Fourier transform. There are abundant examples throughout Answers. (I wrote many of them.)

That result may not be very exciting to view, since ‘X’ is created by returning integer multiples of .