> On Sunday, May 19, 2019 at 4:52:16 AM UTC+12, cl...@claysturner.com wrote:
>> On Tuesday, March 19, 2019 at 12:29:01 PM UTC-4, DSP newbie wrote:
>> > Hi,
>> >
>> > the input signal is sin(t), I multiply it with cos(t) to get the I
>> > component and sin(t) to get the Q component. So then
>> >
>> > I(t)=sin(t)*cos(t)=0.5*sin(2t)
>> > Q(t)=sin(t)^2
>> >
>> > I would expect only the DC component to persist, but instead I have two
>> > quickly wiggling waves. What is missing to shift 1kHz to DC without
>> > introducing the mirror frequencies? Thanks in advance!
>>
>> Since your original signal is real valued, it has a double sided Fourier transform. If you extend your signal to be analytic via a Hilbert transform, then it will now have a one sided Fourier transform.
>>
>> Now when you multiply (complex) by a complex exponential, then you only shift the signal one way. If you shift the signal to baseband, then you will get just a DC signal for your example.
>>
>> Example:
>>
>> Your signal sin(wt)
>>
>> analytic extension is sin(wt)-j cos(wt)
>>
>> mult by complex exponential to base band
>>
>> [ sin(wt)-j cos(wt) ][cos(wt)-j sin(wt)]
>>
>> = sin(wt)cos(wt) -sin(wt)cos(wt) -j[ cos^2(wt) + sin^2(wt)]
>>
>> = 0 - j
>>
>> So there's your constant signal with a -90 degree phase shift.
>>
>> IHTH,
>> Clay
>
> Just as a matter of political correctness, are we allowed to use the
> word Homodyne nowadays? I imagine it would be banned in Europe at
> least.
On Sunday, May 19, 2019 at 4:52:16 AM UTC+12, cl...@claysturner.com wrote:
> On Tuesday, March 19, 2019 at 12:29:01 PM UTC-4, DSP newbie wrote:
> > Hi,
> >
> > the input signal is sin(t), I multiply it with cos(t) to get the I
> > component and sin(t) to get the Q component. So then
> >
> > I(t)=sin(t)*cos(t)=0.5*sin(2t)
> > Q(t)=sin(t)^2
> >
> > I would expect only the DC component to persist, but instead I have two
> > quickly wiggling waves. What is missing to shift 1kHz to DC without
> > introducing the mirror frequencies? Thanks in advance!
>
> Since your original signal is real valued, it has a double sided Fourier transform. If you extend your signal to be analytic via a Hilbert transform, then it will now have a one sided Fourier transform.
>
> Now when you multiply (complex) by a complex exponential, then you only shift the signal one way. If you shift the signal to baseband, then you will get just a DC signal for your example.
>
> Example:
>
> Your signal sin(wt)
>
> analytic extension is sin(wt)-j cos(wt)
>
> mult by complex exponential to base band
>
> [ sin(wt)-j cos(wt) ][cos(wt)-j sin(wt)]
>
> = sin(wt)cos(wt) -sin(wt)cos(wt) -j[ cos^2(wt) + sin^2(wt)]
>
> = 0 - j
>
> So there's your constant signal with a -90 degree phase shift.
>
> IHTH,
> Clay
Just as a matter of political correctness, are we allowed to use the word Homodyne nowadays? I imagine it would be banned in Europe at least.
Reply by ●May 18, 20192019-05-18
On Tuesday, March 19, 2019 at 12:29:01 PM UTC-4, DSP newbie wrote:
> Hi,
>
> the input signal is sin(t), I multiply it with cos(t) to get the I
> component and sin(t) to get the Q component. So then
>
> I(t)=sin(t)*cos(t)=0.5*sin(2t)
> Q(t)=sin(t)^2
>
> I would expect only the DC component to persist, but instead I have two
> quickly wiggling waves. What is missing to shift 1kHz to DC without
> introducing the mirror frequencies? Thanks in advance!
Since your original signal is real valued, it has a double sided Fourier transform. If you extend your signal to be analytic via a Hilbert transform, then it will now have a one sided Fourier transform.
Now when you multiply (complex) by a complex exponential, then you only shift the signal one way. If you shift the signal to baseband, then you will get just a DC signal for your example.
Example:
Your signal sin(wt)
analytic extension is sin(wt)-j cos(wt)
mult by complex exponential to base band
[ sin(wt)-j cos(wt) ][cos(wt)-j sin(wt)]
= sin(wt)cos(wt) -sin(wt)cos(wt) -j[ cos^2(wt) + sin^2(wt)]
= 0 - j
So there's your constant signal with a -90 degree phase shift.
IHTH,
Clay
Reply by Christian Gollwitzer●March 19, 20192019-03-19
Am 19.03.19 um 17:28 schrieb DSP newbie:
> Hi,
>
> the input signal is sin(t), I multiply it with cos(t) to get the I
> component and sin(t) to get the Q component. So then
>
> I(t)=sin(t)*cos(t)=0.5*sin(2t)
> Q(t)=sin(t)^2
>
> I would expect only the DC component to persist, but instead I have two
> quickly wiggling waves.
You need to average it over one period. The I function oscillates around
zero, the Q function around 1/2. so the integral over [0;2*pi] gives 0
for I and pi for Q.
Christian
Reply by Tauno Voipio●March 19, 20192019-03-19
On 19.3.19 18:28, DSP newbie wrote:
> Hi,
>
> the input signal is sin(t), I multiply it with cos(t) to get the I
> component and sin(t) to get the Q component. So then
>
> I(t)=sin(t)*cos(t)=0.5*sin(2t)
> Q(t)=sin(t)^2
>
> I would expect only the DC component to persist, but instead I have two
> quickly wiggling waves. What is missing to shift 1kHz to DC without
> introducing the mirror frequencies? Thanks in advance!
To get the amplitude of the signal, you need to calculate the
length of the IQ phasor. If you do so, you'll see that there
is the double-frequency component left, as it comes from the
image of the non-analytic (single-component) signal.
You do need to make a full complex multiplication to get
rid of the image frequencies.
Google for 'image rejection mixer'.
--
-TV
Reply by DSP newbie●March 19, 20192019-03-19
Hi,
the input signal is sin(t), I multiply it with cos(t) to get the I
component and sin(t) to get the Q component. So then
I(t)=sin(t)*cos(t)=0.5*sin(2t)
Q(t)=sin(t)^2
I would expect only the DC component to persist, but instead I have two
quickly wiggling waves. What is missing to shift 1kHz to DC without
introducing the mirror frequencies? Thanks in advance!