I और Q घटक और QPSK और 4QAM के बीच अंतर

4QAM और QPSK दोनों स्पष्ट रूप से एक ही तरंग उत्पन्न करते हैं, लेकिन क्या वे गणितीय रूप से समान हैं?

QPSK तारामंडल में, 45, 135, 225 और 315 डिग्री पर मानचित्रण बिंदु हैं जबकि 4QAM 0, 90, 180 और 270 पर है?

मैं ऐसे नक्षत्र आरेख के I / Q घटकों को समझने के लिए भी संघर्ष करता हूं। "Inphase" और "quadrature-phase" का वास्तव में क्या मतलब है? क्या वे इस प्रकार के उपयोग के लिए वास्तविक और काल्पनिक भाग को निर्दिष्ट करने का एक और तरीका हैं?

signal-analysis

— chwi
स्रोत

दोनों एक ही हैं। QPSK को QAM का विशेष मामला माना जा सकता है।

— user7234

QPSK और $4$ -QAM तारामंडल दोनों में $45, 135, 225$ और $315$ डिग्री (आपके प्रश्न में नोट टाइपो) पर संकेत बिंदु हैं । वे दो वाहक संकेतों (जिसे inphase और quadrature वाहक कहा जाता है) के आयाम मॉड्यूलेशन (या, यदि आप चाहें, तो चरण मॉड्यूलेशन ) से उत्पन्न होते हैं, जो कि ऑर्थोगोनल हैं (जिसका अर्थ है कि वे चरण में 90 डिग्री से भिन्न हैं। QPSK या $4$ - का कैनन प्रतिनिधित्व) एक प्रतीक अंतराल के दौरान QAM संकेत है

s (t) = (- 1)^{b_{I}} \cos (2 π f_{c} t) - (- 1)^{b_{Q}} \sin (2 π f_{c} t)

$s(t) = (-1)^{b_I}\cos(2\pi f_c t) - (-1)^{b_Q}\sin(2\pi f_c t)$ जहां

\cos (2 π f_{c} t)

$\cos(2\pi f_c t)$ और

- \sin (2 π f_{c} t)

$-\sin(2\pi f_c t)$ कर रहे हैंइंफेजऔरक्षेत्रकलनआवृत्ति पर वाहक के सिग्नल

f_{c}

$f_c$ हर्ट्ज और

b_{I}, b_{Q} \in {0, 1}

$b_I, b_Q \in \{0,1\}$ दो डेटा बिट्स हैं (स्वाभाविक रूप से, इनपेज़ और क्वाडरेचर कैरियर्स पर प्रेषित होने के बाद, इनपेज़ और क्वाडरेचर डेटा बिट्स कहलाते हैं)। सूचना इंफेज वाहक है कि

\cos (2 π f_{c} t)

$\cos(2\pi f_c t)$ हैआयाम

+ 1

$+1$ या

- 1

$-1$ के रूप में इंफेज डेटा बिट है मूल्य के अनुसार

0

$0$ या

1

$1$ , और इसी क्षेत्रकलन वाहक

- \sin (2 π f_{c} t)

$-\sin(2\pi f_c t)$ हैआयाम

+ 1

$+1$ या

- 1

$-1$ क्वाडचर डेटा बिट के अनुसार मान

0

$0$ या

1

$1$ । कुछ लोग इसे चीजों की सामान्य योजना के व्युत्क्रम के रूप में मानते हैं, विशेष रूप से यह कहते हुए कि सकारात्मक आयाम

1

$1$ डेटा बिट्स और

0

$0$ बिट्स के साथ नकारात्मक एम्पलीट्यूड के साथ जुड़ा होना चाहिए । लेकिन अगर हम से इस पर गौर चरण मॉडुलन परिप्रेक्ष्य, एक

0

$0$ बिट साधन वाहक है कि (

\cos (2 π f_{c} t)

$\cos(2\pi f_c t)$ या

- \sin (2 π f_{c} t)

$-\sin(2\pi f_c t)$ जैसा भी मामला हो) के साथ फैलता है में कोई बदलाव नहीं चरणजबकि

1

$1$ डेटा बिट चरण में परिवर्तन करता है (हम इसे चरण विलंब के रूप में सोचेंगे )

180

$180$ डिग्री या

π

$\pi$ रेडियन। दरअसल, QPSK को व्यक्त करने का एक और तरीका है /

4

$4$ -QAM संकेत के रूप में है

s (t) = \cos (2 π f_{c} t - b_{I} π) - \sin (2 π f_{c} t - b_{Q} π)

$s(t) = \cos(2\pi f_c t - b_I\pi) - \sin(2\pi f_c t - b_Q\pi)$ which makes the phase modulation viewpoint very clear. But, regardless of which viewpoint we use, during a symbol interval, the QPSK/

4

$4$ -QAM signal is one of the following four signals:

\sqrt{2} \cos (2 π f_{c} t + \frac{π}{4}), \sqrt{2} \cos (2 π f_{c} t + \frac{3 π}{4}), \sqrt{2} \cos (2 π f_{c} t + \frac{5 π}{4}), \sqrt{2} \cos (2 π f_{c} t + \frac{7 π}{4})

$\sqrt{2}\cos\left(2\pi f_c t + \frac{\pi}{4}\right),\\ \sqrt{2}\cos\left(2\pi f_c t + \frac{3\pi}{4}\right),\\ \sqrt{2}\cos\left(2\pi f_c t + \frac{5\pi}{4}\right),\\ \sqrt{2}\cos\left(2\pi f_c t + \frac{7\pi}{4}\right)$ corresponding to

(b_{I}, b_{Q}) = (0, 0), (1, 0), (1, 1), (0, 1)

$(b_I,b_Q) = (0,0), (1,0), (1,1), (0,1)$ respectively.

Note that the viewpoint taken here is of QPSK as consisting of two BPSK signals on phase-orthogonal carriers. The demodulator thus consists of two BPSK receivers (called the inphase branch and quadrature branch, what else?). An alternative view of QPSK as changing the phase of a single carrier depending on a $4$ -valued symbol is developed a little later.

The QPSK/ $4$ -QAM signal can also be expressed as

s (t) = Re {B \exp (j 2 π f_{c} t)} = Re {[(- 1)^{b_{I}} + j (- 1)^{b_{Q}}] \exp (j 2 π f_{c} t)}

$s(t) = \text{Re}\{B \exp(j2\pi f_c t)\} = \text{Re}\{[(-1)^{b_I}+j(-1)^{b_Q}] \exp(j2\pi f_c t)\}$ where

B

$B$ is the complex-valued baseband symbol taking on values in

{\pm 1 \pm j}

$\{\pm 1 \pm j\}$ and which, when plotted on the complex plane, gives constellation points distant

\sqrt{2}

$\sqrt{2}$ from the origin and at

45, 135, 225

$45, 135, 225$ , and

315

$315$ degrees corresponding to data bits

(b_{I}, b_{Q}) = (0, 0), (1, 0), (1, 1), (0, 1)

$(b_I,b_Q) = (0,0), (1,0), (1,1), (0,1)$ respectively. Note that complementary bit pairs lie diagonally across the circle from each other so that double bit errors are less likely than single bit errors. Note also that the bits naturally occur around the circle in Gray code order; there is no need to massage a given data bit pair

(d_{I}, d_{Q})

$(d_I,d_Q)$ (say

(0, 1)

$(0,1)$ ) from "natural representation" (where it means the integer

2 = d_{I} + 2 d_{Q}

$2 = d_I+2d_Q$ :

d_{I}

$d_I$ is the LSB and

d_{Q}

$d_Q$ the MSB here) to "Gray code representation"

(b_{I}, b_{Q}) = (1, 1)

$(b_I,b_Q) = (1,1)$ of the integer

2

$2$ as some implementations seem to insist on doing. Indeed, such massaging leads to poorer BER performance since the decoded

({\hat{b}}_{I}, {\hat{b}}_{Q})

$(\hat{b}_I,\hat{b}_Q)$ must be ummassaged at the receiver into the decoded data bits

({\hat{d}}_{I}, {\hat{d}}_{Q})

$(\hat{d}_I,\hat{d}_Q)$ making the single channel bit error

(b_{I}, b_{Q}) = (1, 1) \to ({\hat{b}}_{I}, {\hat{b}}_{Q}) = (1, 0)

$(b_I,b_Q) = (1,1) \to (\hat{b}_I,\hat{b}_Q) = (1,0)$ into the double data bit error

(d_{I}, d_{Q}) = (0, 1) \to (b_{I}, b_{Q}) = (1, 1) \to ({\hat{b}}_{I}, {\hat{b}}_{Q}) = (1, 0) \to ({\hat{d}}_{I}, {\hat{d}}_{Q}) = (1, 0) .

${(d_I,d_Q) = (0,1) \to (b_I,b_Q) = (1,1)\to (\hat{b}_I,\hat{b}_Q) = (1,0)}\to (\hat{d}_I,\hat{d}_Q) = (1,0).$

If we delay the four possible signals exhibited above by $45$ degrees or $\pi/4$ radians (subtract $\pi/4$ radians from the argument of the cosinusoid), we get

\sqrt{2} \cos (2 π f_{c} t + \frac{π}{4}) \Rightarrow \sqrt{2} \cos (2 π f_{c} t + 0 \frac{π}{2}) = \sqrt{2} \cos (2 π f_{c} t), \sqrt{2} \cos (2 π f_{c} t + \frac{3 π}{4}) \Rightarrow \sqrt{2} \cos (2 π f_{c} t + 1 \frac{π}{2}) = - \sqrt{2} \sin (2 π f_{c} t), \sqrt{2} \cos (2 π f_{c} t + \frac{5 π}{4}) \Rightarrow \sqrt{2} \cos (2 π f_{c} t + 2 \frac{π}{2}) = - \sqrt{2} \cos (2 π f_{c} t) \sqrt{2} \cos (2 π f_{c} t + \frac{7 π}{4}) \Rightarrow \sqrt{2} \cos (2 π f_{c} t + 3 \frac{π}{2}) = \sqrt{2} \sin (2 π f_{c} t),

$\sqrt{2}\cos\left(2\pi f_c t + \frac{\pi}{4}\right)\Rightarrow \sqrt{2}\cos\left(2\pi f_c t + 0\frac{\pi}{2}\right) = \sqrt{2}\cos(2\pi f_c t),\\ \sqrt{2}\cos\left(2\pi f_c t + \frac{3\pi}{4}\right)\Rightarrow \sqrt{2}\cos\left(2\pi f_c t + 1\frac{\pi}{2}\right) = -\sqrt{2}\sin(2\pi f_c t),\\ \sqrt{2}\cos\left(2\pi f_c t + \frac{5\pi}{4}\right) \Rightarrow \sqrt{2}\cos\left(2\pi f_c t + 2\frac{\pi}{2}\right) = -\sqrt{2}\cos(2\pi f_c t) \\ \sqrt{2}\cos\left(2\pi f_c t + \frac{7\pi}{4}\right)\Rightarrow \sqrt{2}\cos\left(2\pi f_c t + 3\frac{\pi}{2}\right) = \sqrt{2}\sin(2\pi f_c t),\\$ which give the four constellation points at

0, 90, 180, 270

$0,90,180,270$ degrees referred to by the OP. This form gives us another way of viewing QPSK signaling: a single carrier signal whose phase takes on four values depending on the input symbol which takes on values

{0, 1, 2, 3}

$\{0,1,2,3\}$ . We express this in tabular form.

\begin{array}{ccccccc} (b_{I}, b_{Q}) & normal value k & Gray code value ℓ & signal as above & phase-modulated signal \\ (0, 0) & 0 & 0 & \sqrt{2} \cos (2 π f_{c} t) & \sqrt{2} \cos (2 π f_{c} t - 0 \frac{π}{2}) \\ (0, 1) & 1 & 1 & \sqrt{2} \sin (2 π f_{c} t) & \sqrt{2} \cos (2 π f_{c} t - 1 \frac{π}{2}) \\ (1, 1) & 3 & 2 & - \sqrt{2} \cos (2 π f_{c} t) & \sqrt{2} \cos (2 π f_{c} t - 2 \frac{π}{2}) \\ (1, 0) & 2 & 3 & - \sqrt{2} \sin (2 π f_{c} t) & \sqrt{2} \cos (2 π f_{c} t - 3 \frac{π}{2}) \end{array}

$\begin{array}{|c|c|c|c|c|c|c|} \hline (b_I,b_Q) & \text{normal value} ~k & \text{Gray code value} ~\ell & \text{signal as above} &\text{phase-modulated signal}\\ \hline (0,0) & 0 & 0 & \sqrt{2}\cos(2\pi f_c t) & \sqrt{2}\cos\left(2\pi f_c t - 0\frac{\pi}{2}\right)\\ (0,1) & 1 & 1 & \sqrt{2}\sin(2\pi f_c t) & \sqrt{2}\cos\left(2\pi f_c t - 1\frac{\pi}{2}\right)\\ (1,1) & 3 & 2 & -\sqrt{2}\cos(2\pi f_c t) & \sqrt{2}\cos\left(2\pi f_c t - 2\frac{\pi}{2}\right)\\ (1,0) & 2 & 3 & -\sqrt{2}\sin(2\pi f_c t) & \sqrt{2}\cos\left(2\pi f_c t - 3\frac{\pi}{2}\right)\\ \hline \end{array}$ That is, we can regard the QPSK modulator as having input

(b_{I}, b_{Q})

$(b_I,b_Q)$ that it regards as the Gray code representation of the integer

ℓ \in {0, 1, 2, 3}

$\ell \in \{0,1,2,3\}$ and produces the output

\sqrt{2} \cos (2 π f_{c} t - ℓ \frac{π}{2}) .

$\sqrt{2}\cos\left(2\pi f_c t - \ell\frac{\pi}{2}\right).$ In other words, the phase of carrier

\sqrt{2} \cos (2 π f_{c} t)

$\sqrt{2}\cos(2\pi f_c t)$ is modulated (changed from

0

$0$ to

ℓ \frac{π}{2}

$\ell\frac{\pi}{2}$ ) in response to the input

ℓ

$\ell$ .

So how does this work in real life or MATLAB, whichever comes first? If we define a QPSK signal as having value $\sqrt{2}\cos\left(2\pi f_c t - \ell\frac{\pi}{2}\right)$ where the value of $\ell$ is typed in as 0 or 1 or 2 or 3, we will get the QPSK signal described above, but the demodulator will produce the bit pair $(b_I, b_Q)$ and we must remember that the output is $\ell$ in Gray code interpretation, that is, the demodulator output will be $(1,1)$ if $\ell$ happened to have value $2$ , and interpreting output $(1,1)$ as $3$ is a decoding error that is not generally discussed in textbooks!

— Dilip Sarwate
स्रोत

This is the most incredible answer I have ever gotten at SE! Even though I see I have a lot to wrap my mind around, thank you very much! Amazing...

— chwi

My hat's off to Dilip for his fantastic answer. On a purely practical note however, if you were to write a receiver for 4QAM and QPSK, and you have to correct for an arbitrary phase offset, it should be clear that the physical layer receiver for one will work as a physical layer receiver for the other. Also - again, not to diminish Dilip's answer, but the simplest explanation of how IQ can relate to real-valued samples is here

— Dave C

@Dilip Sarwate Excellent answer. Just one doubt, can i assume that QPSK can be achieved by two ways. First one being just amplitude modulating and sending on I and Q channels or second way by only phase modulating the signal by -lpi/2 where l={0,1,2,3}. So you needn't do a combination of both amplitude and phase modulation. Am I right in believing that i need to do both amplitude and phase modulation together to achieve higher orders of QAM like 16-QAM and 64-QAM?

— Karan Talasila

In practice, QPSK is almost universally achieved in just one way: antipodal BPSK on the I and Q carriers, and it results in 4-QAM. You can view it as phase modulation if you like but antipodal BPSK is the same as

2

$2$ -PAM or amplitude modulation and nobody uses a general-purpose

M

$M$ -ary phase modulation circuit (or DSP software subroutine) with

M

$M$ set to

2

$2$ for this purpose. In practice,

2^{2 m}

$2^{2m}$ -QAM is achieved by

2^{m}

$2^m$ -PAM on the I and Q carriers and no phase modulation is used. Note that for

m > 1

$m > 1$ , the PAM cannot be viewed (except by extreme nitpickers) as phase modulation either.

— Dilip Sarwate

@Talasila The A in QAM stands for amplitude.

— Dilip Sarwate