Interactive QPSK Simulation Online

QPSK Modulation Lab

Digital Signal Processing Simulator

System Parameters

Input Bitstream (Dibits)

Carrier Frequency ($f_c$) 5 Hz

Symbol Rate ($R_s$) 1 Baud

Signal Amplitude ($A$) 1.0

Quadrature Phase Shift Keying (QPSK) transmits 2 bits per symbol by shifting the phase of a carrier into one of four states: 45°, 135°, 225°, or 315°.

$$ I_k, Q_k = \text{Map}\{b_{2k}, b_{2k+1}\} \in \left\{ \pm \frac{1}{\sqrt{2}} \right\} $$

The factor \( \tfrac{1}{\sqrt{2}} \) normalizes the QPSK constellation so that each symbol has unit energy (\(I_k^2 + Q_k^2 = 1\)).

$$ I(t) = \sum I_k p(t-kT_s), \quad Q(t) = \sum Q_k p(t-kT_s) $$

$$ \theta_k = \arg(I_k + jQ_k) $$

$$ s(t) = A [ I(t)\cos(2\pi f_c t) - Q(t)\sin(2\pi f_c t) ] $$

Relation Between Mapping and 90° Phase Separation

The mapping assigns each 2-bit combination to one of the four QPSK constellation points. These points are equally spaced around the origin, resulting in a 90° phase separation between adjacent symbols.

Bits	Symbol (I, Q)	Phase
00	(+1, +1)	45°
01	(-1, +1)	135°
11	(-1, -1)	225°
10	(+1, -1)	315°

The phase difference between adjacent symbols is

$$135^\circ - 45^\circ = 90^\circ$$

Similarly,

$$225^\circ - 135^\circ = 90^\circ,\qquad 315^\circ - 225^\circ = 90^\circ$$

Note: The 90° phase separation comes from the positions of the four constellation points. The bit labels (00, 01, 11, 10) simply identify these points and are typically chosen using Gray coding to minimize bit errors.

Why is \( \frac{1}{\sqrt{2}} \) Used in QPSK?

Suppose we map the bits as follows (without normalization):

Bits	I	Q
00	+1	+1
01	-1	+1
11	-1	-1
10	+1	-1

The transmitted QPSK symbol is

$$ s = I + jQ $$

For the symbol (+1, +1):

• \(I = 1\)
• \(Q = 1\)

The symbol energy is

$$ E = I^2 + Q^2 $$

Substituting the values:

$$ E = 1^2 + 1^2 = 2 $$

Therefore, every QPSK symbol has an energy of 2, which is higher than the commonly desired unit energy.

Normalize the Symbol Energy

In digital communications, it is standard practice to normalize the average symbol energy to 1.

To achieve this, divide both the I and Q components by \( \sqrt{2} \):

$$ I = \pm\frac{1}{\sqrt{2}}, \qquad Q = \pm\frac{1}{\sqrt{2}} $$

The symbol energy now becomes

$$ E = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} + \frac{1}{2} = 1 $$

Thus, every QPSK symbol has unit energy. The factor \( \frac{1}{\sqrt{2}} \) is a normalization factor—it does not determine the symbol phase (which remains \(45^\circ\), \(135^\circ\), \(225^\circ\), and \(315^\circ\)).

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