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Adding two cosine waves, result in the standard A*cos(ωt + B) form

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1

1

$begingroup$

Let's say I have the composite wave:

$y(t) = cos(40t) - 0.3 cos(40t - 16)$

This combines two waves of the same frequency but different phases, and the objective here is to add them and present the result in the simplified form, as a single real amplitude multiplied by a single cosine. I have tried complex analysis, via $operatorname{Re}[e^{i40t}(1-0.3e^{-i16})]$, but I still can't get a result out of it. May I ask what I might be missing here?

trigonometry signal-processing

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edited Mar 10 at 14:04

Bernard

123k741116

asked Mar 10 at 13:58

aaaaaaaaaa

85

New contributor

aaaaa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

1

1

$begingroup$

Let's say I have the composite wave:

$y(t) = cos(40t) - 0.3 cos(40t - 16)$

This combines two waves of the same frequency but different phases, and the objective here is to add them and present the result in the simplified form, as a single real amplitude multiplied by a single cosine. I have tried complex analysis, via $operatorname{Re}[e^{i40t}(1-0.3e^{-i16})]$, but I still can't get a result out of it. May I ask what I might be missing here?

trigonometry signal-processing

share|cite|improve this question

edited Mar 10 at 14:04

Bernard

123k741116

asked Mar 10 at 13:58

aaaaaaaaaa

85

New contributor

aaaaa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Let's say I have the composite wave:

$y(t) = cos(40t) - 0.3 cos(40t - 16)$

This combines two waves of the same frequency but different phases, and the objective here is to add them and present the result in the simplified form, as a single real amplitude multiplied by a single cosine. I have tried complex analysis, via $operatorname{Re}[e^{i40t}(1-0.3e^{-i16})]$, but I still can't get a result out of it. May I ask what I might be missing here?

trigonometry signal-processing

share|cite|improve this question

edited Mar 10 at 14:04

Bernard

123k741116

asked Mar 10 at 13:58

aaaaaaaaaa

85

New contributor

aaaaa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 10 at 14:04

Bernard

123k741116

asked Mar 10 at 13:58

aaaaaaaaaa

85

New contributor

aaaaa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 10 at 14:04

Bernard

123k741116

asked Mar 10 at 13:58

aaaaaaaaaa

85

New contributor

aaaaa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 10 at 14:04

Bernard

123k741116

edited Mar 10 at 14:04

Bernard

123k741116

asked Mar 10 at 13:58

aaaaaaaaaa

85

New contributor

aaaaa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 10 at 13:58

aaaaaaaaaa

85

2 Answers
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0

$begingroup$

your working is fine until

$Re[e^{j40t}[1-0.3e^{-j16}]]$ now, remove time varying part ( harmonic part) and focus on it's phasor and write it in Euler form as i.e, $[1-0.3e^{-j16}]=1.28-0.0863j=1.282 e^{-j0.0673}$

since wave is nothing but $=$ Re part of $[ text{phasor}$$times$ $e^{jomega t}]$

so, your wave is $1.282 cos(omega t-0.0673)$ in your case phasor is rotating with frequency 40 radian per second in Anti clock wise direction i.e,put $omega=40$

share|cite|improve this answer

answered Mar 10 at 14:51

Faraday PathakFaraday Pathak

1,165316

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, plus the fact that you calculate the absolute value with the square root of Re^2 + Im^2, and the argument of the exponential with the arctangent of Im/Re - both things that the guy from the other answer directly mentioned. Using complex numbers always simplifies it, so I'm picking your answer as the most straightforward one.
  $endgroup$
  – aaaaa
  Mar 10 at 20:25
  

  
  
  
  

add a comment | 

0

$begingroup$

$$cos{(40t)}-0.3cos{(40t-16)}$$
$$=cos{(40t)}-0.3cos{(16)}cos{(40t)}-0.3sin{(16)}sin{(40t)}$$
$$=(1-0.3cos{(16)})cos{(40t)}-0.3sin{(16)}sin{(40t)}$$
Now we want the above expression in the form
$$R cos{(40t+alpha)}=Rcos{(alpha)}cos{(40t)}-Rsin{(alpha)}sin{(40t)}$$
So, equating coefficients of $cos{(40t)}$ and $sin{(40t)}$ we have
$$Rcos{(alpha)}=1-0.3cos{(16)}$$
$$Rsin{(alpha)}=0.3sin{(16)}$$
Which we can solve simultaneously to give
$$R=sqrt{(1-0.3cos{(16)})^2+(0.3sin{(16)})^2}$$
$$alpha=arctan{bigg(frac{0.3sin{(16)}}{1-0.3cos{(16)}}bigg)}$$
So, the answer is
$$cos{(40t)}-0.3cos{(40t-16)}$$
$$=sqrt{(1-0.3cos{(16)})^2+(0.3sin{(16)})^2} timescos{bigg(40t+arctan{bigg(frac{0.3sin{(16)}}{1-0.3cos{(16)}}bigg)}bigg)}$$
$$approx 1.290192113cos{(40t-0.06699439337)}$$

share|cite|improve this answer

edited Mar 10 at 14:30

answered Mar 10 at 14:24

Peter ForemanPeter Foreman

3,7921216

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add a comment | 

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0

$begingroup$

your working is fine until

$Re[e^{j40t}[1-0.3e^{-j16}]]$ now, remove time varying part ( harmonic part) and focus on it's phasor and write it in Euler form as i.e, $[1-0.3e^{-j16}]=1.28-0.0863j=1.282 e^{-j0.0673}$

since wave is nothing but $=$ Re part of $[ text{phasor}$$times$ $e^{jomega t}]$

so, your wave is $1.282 cos(omega t-0.0673)$ in your case phasor is rotating with frequency 40 radian per second in Anti clock wise direction i.e,put $omega=40$

share|cite|improve this answer

answered Mar 10 at 14:51

Faraday PathakFaraday Pathak

1,165316

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, plus the fact that you calculate the absolute value with the square root of Re^2 + Im^2, and the argument of the exponential with the arctangent of Im/Re - both things that the guy from the other answer directly mentioned. Using complex numbers always simplifies it, so I'm picking your answer as the most straightforward one.
  $endgroup$
  – aaaaa
  Mar 10 at 20:25
  

  
  
  
  

add a comment | 

0

$begingroup$

your working is fine until

$Re[e^{j40t}[1-0.3e^{-j16}]]$ now, remove time varying part ( harmonic part) and focus on it's phasor and write it in Euler form as i.e, $[1-0.3e^{-j16}]=1.28-0.0863j=1.282 e^{-j0.0673}$

since wave is nothing but $=$ Re part of $[ text{phasor}$$times$ $e^{jomega t}]$

so, your wave is $1.282 cos(omega t-0.0673)$ in your case phasor is rotating with frequency 40 radian per second in Anti clock wise direction i.e,put $omega=40$

share|cite|improve this answer

answered Mar 10 at 14:51

Faraday PathakFaraday Pathak

1,165316

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, plus the fact that you calculate the absolute value with the square root of Re^2 + Im^2, and the argument of the exponential with the arctangent of Im/Re - both things that the guy from the other answer directly mentioned. Using complex numbers always simplifies it, so I'm picking your answer as the most straightforward one.
  $endgroup$
  – aaaaa
  Mar 10 at 20:25
  

  
  
  
  

add a comment | 

$begingroup$

your working is fine until

$Re[e^{j40t}[1-0.3e^{-j16}]]$ now, remove time varying part ( harmonic part) and focus on it's phasor and write it in Euler form as i.e, $[1-0.3e^{-j16}]=1.28-0.0863j=1.282 e^{-j0.0673}$

since wave is nothing but $=$ Re part of $[ text{phasor}$$times$ $e^{jomega t}]$

so, your wave is $1.282 cos(omega t-0.0673)$ in your case phasor is rotating with frequency 40 radian per second in Anti clock wise direction i.e,put $omega=40$

share|cite|improve this answer

answered Mar 10 at 14:51

Faraday PathakFaraday Pathak

1,165316

$endgroup$

share|cite|improve this answer

answered Mar 10 at 14:51

Faraday PathakFaraday Pathak

1,165316

answered Mar 10 at 14:51

Faraday PathakFaraday Pathak

1,165316

answered Mar 10 at 14:51

Faraday PathakFaraday Pathak

1,165316

0

$begingroup$

$$cos{(40t)}-0.3cos{(40t-16)}$$
$$=cos{(40t)}-0.3cos{(16)}cos{(40t)}-0.3sin{(16)}sin{(40t)}$$
$$=(1-0.3cos{(16)})cos{(40t)}-0.3sin{(16)}sin{(40t)}$$
Now we want the above expression in the form
$$R cos{(40t+alpha)}=Rcos{(alpha)}cos{(40t)}-Rsin{(alpha)}sin{(40t)}$$
So, equating coefficients of $cos{(40t)}$ and $sin{(40t)}$ we have
$$Rcos{(alpha)}=1-0.3cos{(16)}$$
$$Rsin{(alpha)}=0.3sin{(16)}$$
Which we can solve simultaneously to give
$$R=sqrt{(1-0.3cos{(16)})^2+(0.3sin{(16)})^2}$$
$$alpha=arctan{bigg(frac{0.3sin{(16)}}{1-0.3cos{(16)}}bigg)}$$
So, the answer is
$$cos{(40t)}-0.3cos{(40t-16)}$$
$$=sqrt{(1-0.3cos{(16)})^2+(0.3sin{(16)})^2} timescos{bigg(40t+arctan{bigg(frac{0.3sin{(16)}}{1-0.3cos{(16)}}bigg)}bigg)}$$
$$approx 1.290192113cos{(40t-0.06699439337)}$$

share|cite|improve this answer

edited Mar 10 at 14:30

answered Mar 10 at 14:24

Peter ForemanPeter Foreman

3,7921216

$endgroup$

add a comment | 

0

$begingroup$

$$cos{(40t)}-0.3cos{(40t-16)}$$
$$=cos{(40t)}-0.3cos{(16)}cos{(40t)}-0.3sin{(16)}sin{(40t)}$$
$$=(1-0.3cos{(16)})cos{(40t)}-0.3sin{(16)}sin{(40t)}$$
Now we want the above expression in the form
$$R cos{(40t+alpha)}=Rcos{(alpha)}cos{(40t)}-Rsin{(alpha)}sin{(40t)}$$
So, equating coefficients of $cos{(40t)}$ and $sin{(40t)}$ we have
$$Rcos{(alpha)}=1-0.3cos{(16)}$$
$$Rsin{(alpha)}=0.3sin{(16)}$$
Which we can solve simultaneously to give
$$R=sqrt{(1-0.3cos{(16)})^2+(0.3sin{(16)})^2}$$
$$alpha=arctan{bigg(frac{0.3sin{(16)}}{1-0.3cos{(16)}}bigg)}$$
So, the answer is
$$cos{(40t)}-0.3cos{(40t-16)}$$
$$=sqrt{(1-0.3cos{(16)})^2+(0.3sin{(16)})^2} timescos{bigg(40t+arctan{bigg(frac{0.3sin{(16)}}{1-0.3cos{(16)}}bigg)}bigg)}$$
$$approx 1.290192113cos{(40t-0.06699439337)}$$

share|cite|improve this answer

edited Mar 10 at 14:30

answered Mar 10 at 14:24

Peter ForemanPeter Foreman

3,7921216

$endgroup$

add a comment | 

$begingroup$

$$cos{(40t)}-0.3cos{(40t-16)}$$
$$=cos{(40t)}-0.3cos{(16)}cos{(40t)}-0.3sin{(16)}sin{(40t)}$$
$$=(1-0.3cos{(16)})cos{(40t)}-0.3sin{(16)}sin{(40t)}$$
Now we want the above expression in the form
$$R cos{(40t+alpha)}=Rcos{(alpha)}cos{(40t)}-Rsin{(alpha)}sin{(40t)}$$
So, equating coefficients of $cos{(40t)}$ and $sin{(40t)}$ we have
$$Rcos{(alpha)}=1-0.3cos{(16)}$$
$$Rsin{(alpha)}=0.3sin{(16)}$$
Which we can solve simultaneously to give
$$R=sqrt{(1-0.3cos{(16)})^2+(0.3sin{(16)})^2}$$
$$alpha=arctan{bigg(frac{0.3sin{(16)}}{1-0.3cos{(16)}}bigg)}$$
So, the answer is
$$cos{(40t)}-0.3cos{(40t-16)}$$
$$=sqrt{(1-0.3cos{(16)})^2+(0.3sin{(16)})^2} timescos{bigg(40t+arctan{bigg(frac{0.3sin{(16)}}{1-0.3cos{(16)}}bigg)}bigg)}$$
$$approx 1.290192113cos{(40t-0.06699439337)}$$

share|cite|improve this answer

edited Mar 10 at 14:30

answered Mar 10 at 14:24

Peter ForemanPeter Foreman

3,7921216

$endgroup$

share|cite|improve this answer

edited Mar 10 at 14:30

answered Mar 10 at 14:24

Peter ForemanPeter Foreman

3,7921216

answered Mar 10 at 14:24

Peter ForemanPeter Foreman

3,7921216

answered Mar 10 at 14:24

Peter ForemanPeter Foreman

3,7921216