I have a question:

What is the difference between

$ I \cos(t) - Q \sin(t) $


$ I \cos(t) + Q \sin(t) $

except that they are phase shifted 90 degrees ?

On the internet I can find both modulators with adding quadrature data and substracting quadrature data.

thank you

  • 2
    $\begingroup$ it's actually a 180 degree phase shift (which is what a polarity reversal is) for the quadrature component. $\endgroup$ Sep 14, 2015 at 0:33

2 Answers 2


If we take the complex baseband signal as $S(t) = I(t) + jQ(t)$ (which of course means that we have two separate wires on which the $I(t)$ and $Q(t)$ voltages appear), then we have a nice signal representation for the transmitted signal

$$s(t) = I(t) \cos(2\pi f_c t) - Q(t) \sin(2\pi f_ct) = \operatorname{Re}\left(S(t)e^{j2\pi f_c t}\right).$$ If we continue to fool ourselves by thinking of $S(t)$ having been modulated onto the complex carrier signal (complex sinusoid) $e^{j2\pi f_c t}$, then demodulation is "easy". Simply multiply $S(t)e^{j2\pi f_c t}$ by $e^{-j2\pi f_c t}$ to demodulate the signal and recover $S(t)$. Of course, we don't really have $S(t)e^{j2\pi f_c t}$ in all its complex glory: what we have as the received signal is $\operatorname{Re}\left(S(t)e^{j2\pi f_c t}\right)$. But let us not let that phase us. Let's just multiply $\operatorname{Re}\left(S(t)e^{j2\pi f_c t}\right)$ by $$e^{-j2\pi f_c t} = \cos(2\pi f_c t) - j\sin(2\pi f_c t)$$ which requires two separate multipliers and two separate wires to output the real and imaginary parts. On the real part wire, we get (after ignoring noise, multiplicative factors etc as in MBaz's answer) \begin{align*} s(t)\cos(2\pi f_ct) &= I(t)\cos^2(2\pi f_ct)-Q(t)\sin(2\pi f_ct)\cos(2\pi f_ct)\\ &=I(t)\quad +I(t)\cos(2\pi 2f_ct)-Q(t)\sin(2\pi 2f_ct).\tag{1} \end{align*} while on the imaginary part wire we get \begin{align*} s(t)\left[-\sin(2\pi f_ct)\right] &= -I(t)\cos(2\pi f_ct)\sin(2\pi f_ct)+Q(t)\sin^2(2\pi f_ct)\\ &= Q(t) \quad -I(t)\sin(2\pi 2f_ct)- Q(t)\cos(2\pi 2f_ct).\tag{2} \end{align*} In both $(1)$ and $(2)$, "low-pass" filtering is used to remove the double-frequency terms and leave us with just $I(t)$ and $Q(t)$ on two separate wires, i.e. the complex baseband signal.

So at the transmitter, why multiply $I(t)$ by $\cos(2\pi f_c t)$ and $Q(t)$ by $-\sin(2\pi f_ct)$ and then add to get the transmitted signal $$s(t) = I(t) \cos(2\pi f_c t) - Q(t) \sin(2\pi f_ct)?$$ Why not multiply $Q(t)$ by $+\sin(2\pi f_ct)$ and then add to get $$s(t) = I(t) \cos(2\pi f_c t) + Q(t) \sin(2\pi f_ct)?\tag{3}$$ Sure we can, and now we have the nice representation $\operatorname{Re}\left(S(t)e^{-j2\pi f_c t}\right)$ which works as well except that the company VP will ask you during your presentation why you are using negative frequencies in your carrier signal. At the receiver, we demodulate by multiplying $s(t)$ (as given in $(3)$) by $e^{+j2\pi f_c t}$ so as to get (in the Q branch) \begin{align*} s(t)\sin(2\pi f_ct) &= I(t)\cos(2\pi f_ct)\sin(2\pi f_ct)+Q(t)\sin^2(2\pi f_ct)\\ &= Q(t) \quad +I(t)\sin(2\pi 2f_ct)- Q(t)\cos(2\pi 2f_ct).\tag{4} \end{align*} Note that no complementation/polarity reversal of $Q(t)$ is necessary if we follow a consistent approach and think of everything as complex signals.

  • 1
    $\begingroup$ But let us not let that phase us. I see what you did there! ;-) $\endgroup$
    – Peter K.
    Sep 15, 2015 at 14:33
  • 1
    $\begingroup$ @PeterK. I like to slip in a little thing like that every now and then. Some moderators dislike informality, and would have corrected my spelling. Thank you for not being so prudish. $\endgroup$ Sep 15, 2015 at 14:44
  • $\begingroup$ wow, thank you very much for sharing your knowledge $\endgroup$ Sep 15, 2015 at 23:02

The idea is that you can transmit two signals, $I(t)$ and $Q(t)$, over the same bandwidth and at the same time, and still recover each independently of the other. The math is pretty simple. If the transmitted signal is $s(t)=I(t)\cos(2\pi f_ct)-Q(t)\sin(2\pi f_ct)$, then (ignorning noise, ignoring a factor of 1/2 and assuming coherent reception) the receiver can calculate \begin{align*} r(t)\cos(2\pi f_ct) &= I(t)\cos^2(2\pi f_ct)-Q(t)\sin(2\pi f_ct)\cos(2\pi f_ct)\\ &=I(t)+I(t)\cos(2\pi 2f_ct)-Q(t)\sin(2\pi 2f_ct). \end{align*} After a low-pass filter, the receiver has recovered $I(t)$. Likewise, \begin{align*} r(t)\sin(2\pi f_ct) &= I(t)\cos(2\pi f_ct)\sin(2\pi f_ct)-Q(t)\sin^2(2\pi f_ct)\\ &=I(t)\sin(2\pi 2f_ct)-Q(t)+Q(t)\sin(2\pi 2f_ct). \end{align*} from which $Q(t)$ can be recovered by an low-pass filtering and multiplication by -1.

The identities used here are: $\cos(A)\cos(B)=0.5(\cos(A+B)+\cos(A-B))$, $\cos(A)\sin(B)=0.5(\sin(A+B)-\sin(A-B))$ and $\sin(A)\sin(B)=0.5(\cos(A-B)-\cos(A+B))$.

Of course, you can also define the transmitted signal as $v(t)=I(t)\cos(2\pi f_ct)+Q(t)\sin(2\pi f_ct)$ and avoid the sign inversion when estimating $Q(t)$. The main reason not to, though, is that the complex envelope of $s(t)$ is $I(t)+jQ(t)$. Since most of the time it is preferrable to work with the complex envelope, $s(t)$ is chosen to produce the most convenient expression for it.


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