# Calculation of phase angle − phase difference phase shift from time delay, time of arrival difference, and frequency

Sinusoidal waveforms with identical frequency is capable of having a phase difference.
When there is a phase shift (phase difference) or phase delay with the φ (Greek letter Phi) in degrees it must be described between the particular (sine waves) this presents itself. Therefore, for instance, a phase shift could be between the left and right stereo channel signals, between input and output signal, between voltage and current, or between sound pressure p and velocity of the air particles v.

 One full cycle of the wave involves an "angular" displacement of 2 π radians.                The phase φ can be understood as the angle of a section of a signal, described in angular degrees and offers a reference to the benchmark value of the whole signal. With regard to periodic signals it is the whole phase angle of 360 degrees and a time period corresponding to the period duration. A common question: What is the frequency and the phase angle of a sine waveform? Can "a single" signal genuinely possess a phase? A couple of "in-phase" waves employ a phase (angle) of φ = 0 degrees. When the frequency = 0 Hz, in that case you cannot find any AC voltage - that's only DC. Subsequently you will have absolutely no phase angle existing.
What has time delay to do with phase angle?
 Frequency f Hz Time delay Δ t ms ↓ Phase difference φ in degrees ° or deg φ in radians rad c = 343 m/s at 20°   wavelength λ m
 Relationship between phase angle φ° in degrees (deg), the time delay Δ t and the frequency f is: Phase angle (deg)    (Time shift) Time difference     Frequency    λ = c / f  and  c = 343 m/s at 20°C. Relationship between phase angle φ in radians (rad), the time shift or time delay Δ t, and the frequency f is: Phase angle (rad)    "Bogen" means "radians". (Time shift) Time difference Frequency    Time = path length / speed of sound

The time difference (duration) of sound per meter
Effect of temperature on the time difference Δ t
Dependence of the only on the
 Temperature of air in °C Speed of sound c in m/s Time per 1 m Δ t in ms/m +40 354.9 2.818 +35 352.0 2.840 +30 349.1 2.864 +25 346.2 2.888 +20 343.2 2.912 +15 340.3 2.937 +10 337.3 2.963 +5 334.3 2.990 ±0 331.3 3.017 −5 328.2 3.044 −10 325.2 3.073 −15 322.0 3.103 −20 318.8 3.134 −25 315.7 3.165

 Sound engineers take usually the rule of thumb:  For the distance of r = 1 m the sound needs about t = 3 ms in air.  Δ t = r / c and r = Δ t × c       Speed of sound c = 343 m/s at 20°C.

For time delays that is fixed having Δ t = 0.5 ms we derive

the below given phase shift φ° (deg) of the signal:

 Phase difference φ° (deg) Phase difference φBogen (rad) Frequency  f Wavelength λ = c / f 360° 2 π = 6.283185307 2000 Hz 0.171 m 180° π = 3.141592654 1000 Hz 0.343 m 90° π / 2 = 1.570796327 500 Hz 0.686 m 45° π / 4 = 0.785398163 250 Hz 1.372 m 22.5° π / 8 = 0.392699081 125 Hz 2.744 m 11.25° π /16= 0.196349540 62.5 Hz 5.488 m

Phase angle: φ° = 360 × f × Δ t
For time-based Left right stereophonic Δ t = a × sin α / c
Frequency f = φ° / 360 × Δ t

Phase angle (deg) φ = time delay Δ t × frequency f × 360
If we consider the time difference Δ t = path, length a / speed of sound c, then we find
Phase difference φ° = path length a × frequency f × 360 / speed of sound c

You can try entering two values, and get the third value through the following calculator software.

 Phase angle (deg) φ ° (Time shift) time delay Δ t ms Frequency f Hz
 John William Strutt, 3rd Lord Rayleigh, in the year 1907 proved this theory that helps us to understand the phenomenon of "natural hearing" for humans. It is the simple recognition that the interaural period of arrival differences ITD are crucial in frequencies which may be under 800 Hz as with the the humans , whereas from frequencies over 1600 Hz solely the interaural level differences ILD are usually productive. Between the ears the highest level of delay ranges upto 0.63 ms. Phase differences for specific frequencies could be determined. Phase shifter circuit for phase angles from φ = 0° to 180°
Voltage vectors of the phase shifter

For R = 0 ohm is VOUT = VIN. The output should not be attached with a load having a low impedance.

It is possible shifting single pure frequencies (sine waves), but that may not be impossible for music programs which is given below

Two sine voltages - phase shifted: φ = 45°

Conditions for distortion-free transmission

From Schoeps - Joerg Wuttke: "Mikrofonbuch" - Chapter 7

 Although the necessity for a consistent frequency response is apparent, the "linear" phase demands rather more clarification. You will find technicians who expect to have the perfect phase as a constant much like the amplitude effect. Which is not accurate. At first, the phase commences at 0° since the smallest frequency comes to an end on 0 Hz, at DC. (You cannot find any phase angle between DC voltages). Throughout certain frequency a phase angle will be with no significance, when the phase angle is just two times as big with regards to double frequency, and 3 x as big in triplicate, etc.
(About Comb Filtering, Phase Shift and Polarity Reversal)

 Electronic counterpart of the movement of a signal and its retarded iteration, re-merged into a single signal. In the circumstance we will be investigating, the delay range carries a delay of 1 millisecond, the quantities of both the initial and delayed signals entering the mixer will be identical, and the signal can be a 1 kHz sine wave.

If we consider s sine wave wth 1500 Hz. frequency (period T = 0.667 ms) and its delayed iteration, at 1 ms delay. The mixed signal produced is going to be a signal without any amplitude, or perhaps a total termination of signal.
If we consider s sine wave wth 1500 Hz. frequency (period T = 0.667 ms) and its delayed iteration, at 1 ms delay. The mixed signal produced is going to be a signal without any amplitude, or perhaps a total termination of signal.

The phase shift for every frequency having a 1 millisecond delay the oblicuo line signifies the growing phase shift as a function of frequency. Remember that we are able to imagine 540° to be appropriately similar to 180°. Ø (phi) = phase shift, phase shifting, phase difference, displacement of phase, phase lag, phase angle are all frequently not necessarily accurate as applied as:pol-rev = polarity reversal.

Polarity and phase in many cases are applied as though they signify the same principle. However, they may not be.
The "phase reverse button" is not going to modify the phase. This flips the polarity.

Polarity flipping isn't a phase shift.

Polarity reversal (or Pol-Rev) is often a expression that may be usually mistaken for phase Ø (phi), however consists of no phase shift or time delay. Polarity reversal takes place once we "change the polarity" of the amplitude principles of a signal. Within the analog likeness this can be carried out using an inverting amplifier, a transformer, or in a balanced line by merely changing connections among pins 2 and 3 (XLR plug) for one end of
the cable. Within the digital sphere, it is carried out simply by modifying all pluses to minuses and the other way round in the audio-signal data approach.

Two sawtooth oscillations
 Uppermost: the original signal a/b (saw tooth) Center: the 180° phase shifted signal as T/2 time shifted sawtooth Lowermost: the b/a-polarity reversed (inverted) signal, mirrored over the time axis

Evidently can be found that reversed polarity cannot be very much like out of phase. It is concerning the a lot talked about topic: "Phase shift vs. inverting a signal" and "phase-shift vs. time shift of a signal." The concept of a phase shift is apparently identified only for mono frequency sine signals and the phase shift angle is clearly outlined exclusively for sinusoidal proportions.

The typical Ø (phi)-button is only a polarity changer
There is absolutely no phase shifting

 Be aware: Time, frequency and phase are supposed to be in close proximity with each other. The peak of the amplitude does not have any effect on these variables.
The Angular Frequency is ω = 2π × f
 In the following equation we can see: y = 50 sin (5000 t)Calculate the frequency and the amplitude.Solution: The amplitude is 50 and ω = 5000. Therefore the frequency will be f = 1/T = ω / 2 π = 795.77 Hz.
 To make use of the calculator, you just have to enter a value. The calculator is designed to function in both directions of the ↔ sign.
 Frequency f Hz ↔ Angular Frequency ω rad/s ω = 2π × f                              f = ω / 2π